# Problem 2.5b1

A First Course in String Theory

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2.5 Constructing simple orbifolds

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The wikipedia page “Fundamental polygon”, specifically the subsection entitled “group generators”, has a serious mathematical error. You cannot derive a presentation for the fundamental group from the fundamental polygon using the side labels in the manner described on that page (and which you have copied), unless all of the vertices of the polygon are identified to the same point. In the picture you provided and which can be seen on that page, one opposite pair of vertices of the square is identified to one point on the sphere, the other opposite pair of vertices is identified to a different point on the sphere.

There is still a way to derive a presentation for the fundamental group from a fundamental polygon, but it is not the way described on the wikipedia page. In the sphere example of your question, you have to ignore one of the two letters $\displaystyle{A}$, $\displaystyle{B}$, keeping only the other letter. For example, ignoring $\displaystyle{A}$ and keeping $\displaystyle{B}$, you get a presentation $\displaystyle{ \langle B \mid B B^{-1} = 1 \rangle }$, which is a presentation of the trivial group. The way you tell which to ignore and which to keep is by taking the quotient of the boundary of the polygon which is a graph with vertices and edges, choosing a maximal tree in that graph, ignoring all edge labels in the maximal tree, and keeping all edge labels not in the maximal tree.

On that wikipedia page, the Klein bottle and the torus examples are correct and you do not have to ignore any edge labels: all vertices are identified to a single point and the maximal tree is just a point. The sphere and the projective plane examples are incorrect: the four vertices are identified to two separate points, the maximal tree has one edge, and you have to ignore one edge label. The example of a hexagon fundamental domain for the torus is also incorrect: the six vertices are identified to two separate points, the maximal tree has one edge, and you have to ignore one edge label.

edited Jul 23 ’14 at 17:17

answered Jul 23 ’14 at 17:11

Lee Mosher

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yes, i thought that the fundamental polygon is this quotient space. – user159356 Jul 23 ’14 at 17:28

That’s backward: in your example, the sphere is the quotient space of the fundamental polygon, not the other way around. – Lee Mosher Jul 23 ’14 at 17:30

— Mathematics Stack Exchange

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2021.01.18 Monday ACHK

# Problem 2.5a

A First Course in String Theory

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2.5 Constructing simple orbifolds

(a) Consider a circle $\displaystyle{S^1}$, presented as the real line with the identification $\displaystyle{x \sim x + 2}$. Choose $\displaystyle{-1 < x \le 1}$ as the fundamental domain. The circle is the space $\displaystyle{-1 < x \le 1}$ with points $\displaystyle{x = \pm 1}$ identified. The orbifold $\displaystyle{S^1/\mathbb{Z}_2}$ is defined by imposing the (so-called) $\displaystyle{\mathbb{Z}_2}$ identification $\displaystyle{x \sim -x}$. Describe the action of this identification on the circle. Show that there are two points on the circle that are left fixed by the $\displaystyle{\mathbb{Z}_2}$ action. Find a fundamental domain for the two identifications. Describe the orbifold $\displaystyle{S^1/\mathbb{Z}_2}$ in simple terms.

~~~

Put point $\displaystyle{x=0}$ and point $\displaystyle{x=1}$ on the positions that they can form a horizontal diameter.

Then the action is a reflection of the lower semi-circle through the horizontal diameter to the upper semi-circle.

Point $\displaystyle{x=0}$ and point $\displaystyle{x=1}$ are the two fixed points.

A possible fundamental domain is $\displaystyle{0 \le x \le 1}$.

If a variable point $\displaystyle{x}$ moves from 0 to 1 and then keeps going, that point will actually go back and forth between 0 and 1.

— Me@2020-12-31 04:43:07 PM

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2021.01.01 Friday (c) All rights reserved by ACHK

# Problem 2.4

A First Course in String Theory

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2.4 Lorentz transformations as matrices

A matrix L that satisfies (2.46) is a Lorentz transformation. Show the following.

(b) If $\displaystyle{L}$ is a Lorentz transformation so is the inverse matrix $\displaystyle{L^{-1}}$.

(c) If $\displaystyle{L}$ is a Lorentz transformation so is the transpose matrix $\displaystyle{L^{T}}$.

~~~

(b)

\displaystyle{ \begin{aligned} (\mathbf{A}^\mathrm{T})^{-1} &= (\mathbf{A}^{-1})^\mathrm{T} \\ L^T \eta L &= \eta \\ \eta &= [L^T]^{-1} \eta L^{-1} \\ [L^T]^{-1} \eta L^{-1} &= \eta \\ [L^{-1}]^T \eta L^{-1} &= \eta \\ \end{aligned}}

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(c)

\displaystyle{ \begin{aligned} L^T \eta L &= \eta \\ (L^T \eta L)^{-1} &= (\eta)^{-1} \\ L^{-1} \eta^{-1} (L^T)^{-1} &= \eta \\ L^{-1} \eta (L^T)^{-1} &= \eta \\ \eta &= L \eta L^T \\ L \eta L^T &= \eta \\ \end{aligned}}

— Me@2020-12-21 04:24:33 PM

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2020.12.21 Monday (c) All rights reserved by ACHK

# Problem 2.3b5

A First Course in String Theory

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2.3 Lorentz transformations, derivatives, and quantum operators.

(b) Show that the objects $\displaystyle{\frac{\partial}{\partial x^\mu}}$ transform under Lorentz transformations in the same way as the $\displaystyle{a_\mu}$ considered in (a) do. Thus, partial derivatives with respect to conventional upper-index coordinates $\displaystyle{x^\mu}$ behave as a four-vector with lower indices – as reflected by writing it as $\displaystyle{\partial_\mu}$.

~~~

Denoting $\displaystyle{ \eta_{\mu \rho} L^\rho_{~\sigma} \eta^{\nu \sigma}}$ as $\displaystyle{L^{~\nu}_{\mu}}$ is misleading, because that presupposes that $\displaystyle{ \eta_{\mu \rho} L^\rho_{~\sigma} \eta^{\nu \sigma}}$ is directly related to the matrix $\displaystyle{L}$.

To avoid this bug, instead, we denote $\displaystyle{ \eta_{\mu \rho} L^\rho_{~\sigma} \eta^{\nu \sigma}}$ as $\displaystyle{M ^\nu_{~\mu}}$. So

\displaystyle{ \begin{aligned} (x')^\mu &= L^\mu_{~\nu} x^\nu \\ (x')^\mu (x')_\mu &= \left( L^\mu_{~\nu} x^\nu \right) \left( \eta_{\mu \rho} L^\rho_{~\sigma} \eta^{\beta \sigma} x_\beta \right) \\ (x')^\mu (x')_\mu &= \left( L^\mu_{~\nu} x^\nu \right) \left( M^{\beta}_{~\mu} x_\beta \right) \\ x^\mu x_\mu &= \left( L^\mu_{~\nu} x^\nu \right) \left( M^{\beta}_{~\mu} x_\beta \right) \\ \end{aligned}}

\displaystyle{ \begin{aligned} \nu \neq \mu:&~~~~~~\sum_{\mu = 0}^4 L^\mu_{~\nu} M^{\beta}_{~\mu} &= 0 \\ \nu = \mu:&~~~~~~\sum_{\mu = 0}^4 L^\mu_{~\nu} M^{\beta}_{~\mu} &= 1 \\ \end{aligned}}

Using the Kronecker Delta and Einstein summation notation, we have

\displaystyle{ \begin{aligned} L^\mu_{~\nu} M^{\beta}_{~\mu} &= M^{\beta}_{~\mu} L^\mu_{~\nu} \\ &= \delta^{\beta}_{~\nu} \\ \end{aligned}}

So

\displaystyle{ \begin{aligned} \sum_{\mu=0}^{4} L^\mu_{~\nu} M^{\beta}_{~\mu} &= \delta^{\beta}_{~\nu} \\ \end{aligned}}

\displaystyle{ \begin{aligned} M^{\beta}_{~\mu} &= [L^{-1}]^{\beta}_{~\mu} \\ \end{aligned}}

In other words,

\displaystyle{ \begin{aligned} \eta_{\mu \rho} L^\rho_{~\sigma} \eta^{\nu \sigma} &= [L^{-1}]^{\beta}_{~\mu} \\ \end{aligned}}

— Me@2020-11-23 04:27:13 PM

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One defines (as a matter of notation),

${\displaystyle {\Lambda _{\nu }}^{\mu }\equiv {\left(\Lambda ^{-1}\right)^{\mu }}_{\nu },}$

and may in this notation write

${\displaystyle {A'}_{\nu }={\Lambda _{\nu }}^{\mu }A_{\mu }.}$

Now for a subtlety. The implied summation on the right hand side of

${\displaystyle {A'}_{\nu }={\Lambda _{\nu }}^{\mu }A_{\mu }={\left(\Lambda ^{-1}\right)^{\mu }}_{\nu }A_{\mu }}$

is running over a row index of the matrix representing $\displaystyle{\Lambda^{-1}}$. Thus, in terms of matrices, this transformation should be thought of as the inverse transpose of $\displaystyle{\Lambda}$ acting on the column vector $\displaystyle{A_\mu}$. That is, in pure matrix notation,

${\displaystyle A'=\left(\Lambda ^{-1}\right)^{\mathrm {T} }A.}$

— Wikipedia on Lorentz transformation

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So

\displaystyle{ \begin{aligned} M^{\beta}_{~\mu} &= [L^{-1}]^{\beta}_{~\mu} \\ \end{aligned}}

In other words,

\displaystyle{ \begin{aligned} \eta_{\mu \rho} L^\rho_{~\sigma} \eta^{\beta \sigma} &= [L^{-1}]^{\beta}_{~\mu} \\ \end{aligned}}

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Denote $\displaystyle{[L^{-1}]^{\beta}_{~\mu}}$ as

\displaystyle{ \begin{aligned} N^{~\beta}_{\mu} \\ \end{aligned}}

In other words,

\displaystyle{ \begin{aligned} N^{~\beta}_{\mu} &= M^{\beta}_{~\mu} \\ [N^T] &= [M] \\ \end{aligned}}

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The Lorentz transformation:

\displaystyle{ \begin{aligned} (x')^\mu &= L^\mu_{~\nu} x^\nu \\ (x')_\mu &= \eta_{\mu \rho} L^\rho_{~\sigma} \eta^{\beta \sigma} x_\beta \\ \end{aligned}}

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\displaystyle{ \begin{aligned} (x')^\mu &= L^\mu_{~\nu} x^\nu \\ (x')_\mu &= N^{~\nu}_{\mu} x_\nu \\ \end{aligned}}

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\displaystyle{ \begin{aligned} x^\mu &= [L^{-1}]^\mu_{~\nu} (x')^\nu \\ (x')_\mu &= M^{\nu}_{~\mu} x_\nu \\ \end{aligned}}

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\displaystyle{ \begin{aligned} x^\mu &= [L^{-1}]^\mu_{~\nu} (x')^\nu \\ (x')_\mu &= [L^{-1}]^{\nu}_{~\mu} x_\nu \\ \end{aligned}}

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\displaystyle{ \begin{aligned} \frac{\partial}{\partial (x')^\mu} &= \frac{\partial x^\nu}{\partial (x')^\mu} \frac{\partial}{\partial x^\nu} \\ &= \frac{\partial x^0}{\partial (x')^\mu} \frac{\partial}{\partial x^0} + \frac{\partial x^1}{\partial (x')^\mu} \frac{\partial}{\partial x^1} + \frac{\partial x^2}{\partial (x')^\mu} \frac{\partial}{\partial x^2} + \frac{\partial x^3}{\partial (x')^\mu} \frac{\partial}{\partial x^3} \\ \end{aligned}}

Now we consider $\displaystyle{f}$ as a function of $\displaystyle{x^{\mu}}$‘s:

$\displaystyle{f(x^0, x^1, x^2, x^3)}$

Since $\displaystyle{x^{\mu}}$‘s and $\displaystyle{(x')^{\mu}}$‘s are related by Lorentz transform, $\displaystyle{f}$ is also a function of $\displaystyle{(x')^{\mu}}$‘s, although indirectly.

$\displaystyle{f(x^0((x')^0, (x')^1, (x')^2, (x')^3), x^1((x')^0, ...), x^2((x')^0, ...), x^3((x')^0, ...))}$

For notational simplicity, we write $\displaystyle{f}$ as

$\displaystyle{f(x^\alpha((x')^\beta))}$

Since $\displaystyle{f}$ is a function of $\displaystyle{(x')^{\mu}}$‘s, we can differentiate it with respect to $\displaystyle{(x')^{\mu}}$‘s.

\displaystyle{ \begin{aligned} \frac{\partial}{\partial (x')^\mu} f(x^\alpha((x')^\beta))) &= \sum_{\nu = 0}^4 \frac{\partial x^\nu}{\partial (x')^\mu} \frac{\partial}{\partial x^\nu} f(x^\alpha) \\ \end{aligned}}

Since

\displaystyle{ \begin{aligned} x^\nu &= [L^{-1}]^\nu_{~\beta} (x')^\beta \\ \end{aligned}},

\displaystyle{ \begin{aligned} \frac{\partial f}{\partial (x')^\mu} &= \sum_{\nu = 0}^4 \frac{\partial}{\partial (x')^\mu} \left[ \sum_{\beta = 0}^4 [L^{-1}]^\nu_{~\beta} (x')^\beta \right] \frac{\partial f}{\partial x^\nu} \\ &= \sum_{\nu = 0}^4 \sum_{\beta = 0}^4 [L^{-1}]^\nu_{~\beta} \frac{\partial (x')^\beta}{\partial (x')^\mu} \frac{\partial f}{\partial x^\nu} \\ &= \sum_{\nu = 0}^4 \sum_{\beta = 0}^4 [L^{-1}]^\nu_{~\beta} \delta^\beta_\mu \frac{\partial f}{\partial x^\nu} \\ &= \sum_{\nu = 0}^4 [L^{-1}]^\nu_{~\mu} \frac{\partial f}{\partial x^\nu} \\ &= [L^{-1}]^\nu_{~\mu} \frac{\partial f}{\partial x^\nu} \\ \end{aligned}}

Therefore,

\displaystyle{ \begin{aligned} \frac{\partial}{\partial (x')^\mu} &= [L^{-1}]^\nu_{~\mu} \frac{\partial}{\partial x^\nu} \\ \end{aligned}}

It is the same as the Lorentz transform for covariant vectors:

\displaystyle{ \begin{aligned} (x')_\mu &= [L^{-1}]^{\nu}_{~\mu} x_\nu \\ \end{aligned}}

— Me@2020-11-23 04:27:13 PM

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2020.11.24 Tuesday (c) All rights reserved by ACHK

# Kronecker delta in tensor component form

Problem 2.3b4

A First Course in String Theory

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Continue the previous calculation:

\displaystyle{ \begin{aligned} \nu \neq \mu:&~~~~~~\sum_{\mu = 0}^4 L^\mu_{~\nu} M^{\beta}_{~\mu} &= 0 \\ \nu = \mu:&~~~~~~\sum_{\mu = 0}^4 L^\mu_{~\nu} M^{\beta}_{~\mu} &= 1 \\ \end{aligned}}

The two cases can be grouped into one, by replacing the right hand sides with the Kronecker delta. However, there are 4 possible forms and I am not sure which one should be used.

$\displaystyle{\delta^i_{~j}}$
$\displaystyle{\delta_i^{~j}}$
$\displaystyle{\delta^{ij}}$
$\displaystyle{\delta_{ij}}$

So I do a little research on Kronecker delta in this post.

— Me@2020-10-21 03:40:36 PM

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The inverse Lorentz transformation should satisfy $\displaystyle{\left( \Lambda^{-1} \right)^\beta_{~\mu} \Lambda^\mu_{~\nu} = \delta^\beta_{~\nu}}$, where $\displaystyle{\delta^\beta_{~\nu} \equiv \text{diag}(1,1,1,1)}$ is the Kronecker delta. Then, multiply by the inverse on both sides of Eq. 4 to find

\displaystyle{ \begin{aligned} \left( \Lambda^{-1} \right)^\beta_{~\mu} \left( \Delta x' \right)^\mu &= \delta^\beta_{~\nu} \Delta x^\nu \\ &= \Delta x^\beta \\ \end{aligned}}

(6)

The inverse $\displaystyle{\left( \Lambda^{-1} \right)^\beta_{~\mu}}$ is also written as $\displaystyle{\Lambda_\mu^{~\beta}}$. The notation is as follows: the left index denotes a row while the right index denotes a column, while the top index denotes the frame we’re transforming to and the bottom index denotes the frame we’re transforming from. Then, the operation $\displaystyle{\Lambda_\mu^{~\beta} \Lambda^\mu_{~\nu}}$ means sum over the index $\displaystyle{\mu}$ which lives in the primed frame, leaving unprimed indices $\displaystyle{\beta}$ and $\displaystyle{\nu}$ (so that the RHS of Eq. 6 is unprimed as it should be), where the sum is over a row of $\displaystyle{\Lambda_\mu^{~\beta}}$ and a column of $\displaystyle{\Lambda_{~\nu}^\mu}$ which is precisely the operation of matrix multiplication.

— Lorentz tensor redux

— Emily Nardoni

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This one is WRONG:

$\displaystyle{(\Lambda^T)^{\mu}{}_{\nu} = \Lambda_{\nu}{}^{\mu}}$

This one is RIGHT:

$\displaystyle{(\Lambda^T)_{\nu}{}^{\mu} ~:=~ \Lambda^{\mu}{}_{\nu}}$

— Me@2020-10-23 06:30:57 PM

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1. $\displaystyle{(\Lambda^T)_{\nu}{}^{\mu} ~:=~\Lambda^{\mu}{}_{\nu}}$

2. [Kronecker delta] is invariant in all coordinate systems, and hence it is an isotropic tensor.

3. Covariant, contravariant and mixed type of this tensor are the same, that is

$\displaystyle{\delta^i_{~j} = \delta_i^{~j} = \delta^{ij} = \delta_{ij}}$

— Introduction to Tensor Calculus

— Taha Sochi

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Raising and then lowering the same index (or conversely) are inverse operations, which is reflected in the covariant and contravariant metric tensors being inverse to each other:

${\displaystyle g^{ij}g_{jk}=g_{kj}g^{ji}={\delta ^{i}}_{k}={\delta _{k}}^{i}}$

where $\displaystyle{\delta^i_{~k}}$ is the Kronecker delta or identity matrix. Since there are different choices of metric with different metric signatures (signs along the diagonal elements, i.e. tensor components with equal indices), the name and signature is usually indicated to prevent confusion.

— Wikipedia on Raising and lowering indices

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So

${\displaystyle g^{ij}g_{jk}={\delta ^{i}}_{k}}$

and

${\displaystyle g_{kj}g^{ji}={\delta _{k}}^{i}}$

— Me@2020-10-19 05:21:49 PM

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$\displaystyle{ T_{i}^{\; j} = \boldsymbol{T}(\boldsymbol{e}_i,\boldsymbol{e}^j) }$ and $\displaystyle{T_{j}^{\; i} = \boldsymbol{T}(\boldsymbol{e}_j,\boldsymbol{e}^i) }$ are both 1-covariant 2-contravariant coordinates of T. The only difference between them is the notation used for sub- and superscripts;

$\displaystyle{ T^{i}_{\; j} = \boldsymbol{T}(\boldsymbol{e}^i,\boldsymbol{e}_j) }$ and $\displaystyle{ T^{j}_{\; i} = \boldsymbol{T}(\boldsymbol{e}^j,\boldsymbol{e}_i) }$ are both 1-contravariant 2-covariant coordinates of T. The only difference between them is the notation used for sub- and superscripts.

— edited Oct 11 ’17 at 14:14

— answered Oct 11 ’17 at 10:58

— EditPiAf

— Tensor Notation Upper and Lower Indices

— Physics StackExchange

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Rather, the dual basis one-forms are defined by imposing the following
16 requirements at each spacetime point:

$\displaystyle{\langle \tilde{e}^\mu \mathbf{x}, \vec e_\nu \mathbf{x} \rangle = \delta^{\mu}_{~\nu}}$

is the Kronecker delta, $\displaystyle{\delta^{\mu}_{~\nu} = 1}$ if $\displaystyle{\mu = \nu}$ and $\displaystyle{\delta^{\mu}_{~\nu} = 0}$ otherwise, with the same values for each spacetime point. (We must always distinguish subscripts from superscripts; the Kronecker delta always has one of each.)

— Introduction to Tensor Calculus for General Relativity

— Edmund Bertschinger

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However, since $\displaystyle{\delta_{~b}^a}$ is a tensor, we can raise or lower its indices using the metric tensor in the usual way. That is, we can get a version of $\displaystyle{\delta}$ with both indices raised or lowered, as follows:

[$\displaystyle{\delta^{ab} = \delta^a_{~c} g^{cb} = g^{ab}}$]

$\displaystyle{\delta_{ab} = g_{ac} \delta^c_{~b} = g_{ab}}$

In this sense, $\displaystyle{\delta^{ab}}$ and $\displaystyle{\delta_{ab}}$ are the upper and lower versions of the metric tensor. However, they can’t really be considered versions of the Kronecker delta any more, as they don’t necessarily satisfy [0 when $i \ne j$ and 1 when $i = j$]. In other words, the only version of $\delta$ that is both a Kronecker delta and a tensor is the version with one upper and one lower index: $\delta^a_{~b}$ [or $\delta^{~a}_{b}$].

— Kronecker Delta as a tensor

— physicspages

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Continue the calculation for the Problem 2.3b:

Denoting $\displaystyle{ \eta_{\mu \rho} L^\rho_{~\sigma} \eta^{\nu \sigma}}$ as $\displaystyle{L^{~\nu}_{\mu}}$ is misleading, because that presupposes that $\displaystyle{ \eta_{\mu \rho} L^\rho_{~\sigma} \eta^{\nu \sigma}}$ is directly related to the matrix $\displaystyle{L}$.

To avoid this bug, instead, we denote $\displaystyle{ \eta_{\mu \rho} L^\rho_{~\sigma} \eta^{\nu \sigma}}$ as $\displaystyle{M ^\nu_{~\mu}}$. So

\displaystyle{ \begin{aligned} (x')^\mu &= L^\mu_{~\nu} x^\nu \\ (x')^\mu (x')_\mu &= \left( L^\mu_{~\nu} x^\nu \right) \left( \eta_{\mu \rho} L^\rho_{~\sigma} \eta^{\beta \sigma} x_\beta \right) \\ (x')^\mu (x')_\mu &= \left( L^\mu_{~\nu} x^\nu \right) \left( M^{\beta}_{~\mu} x_\beta \right) \\ x^\mu x_\mu &= \left( L^\mu_{~\nu} x^\nu \right) \left( M^{\beta}_{~\mu} x_\beta \right) \\ \end{aligned}}

\displaystyle{ \begin{aligned} \nu \neq \mu:&~~~~~~\sum_{\mu = 0}^4 L^\mu_{~\nu} M^{\beta}_{~\mu} &= 0 \\ \nu = \mu:&~~~~~~\sum_{\mu = 0}^4 L^\mu_{~\nu} M^{\beta}_{~\mu} &= 1 \\ \end{aligned}}

Using the Kronecker Delta and Einstein summation notation, we have

\displaystyle{ \begin{aligned} L^\mu_{~\nu} M^{\beta}_{~\mu} &= M^{\beta}_{~\mu} L^\mu_{~\nu} \\ &= \delta^{\beta}_{~\nu} \\ \end{aligned}}

Note: After tensor contraction, the remaining left index should be kept on the left and the remaining right on the right.

— Me@2020-10-20 03:49:09 PM

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2020.10.21 Wednesday (c) All rights reserved by ACHK

# Problem 2.3b3

Now we lower the indices, by expressing the upper-index coordinates (contravariant components) by lower-index coordinates (covariant components), in order to find the Lorentz transformation for the covariant components:

\displaystyle{ \begin{aligned} (x')^\mu &= L^\mu_{~\nu} x^\nu \\ \eta^{\rho \mu} (x')_\rho &= L^\mu_{~\nu} \eta^{\sigma \nu} x_\sigma \\ \sum_\rho \eta^{\rho \mu} (x')_\rho &= \sum_\sigma \sum_\nu L^\mu_{~\nu} \eta^{\sigma \nu} x_\sigma \\ \end{aligned}}

After raising the indices, we lower the indices again:

\displaystyle{ \begin{aligned} \eta_{\alpha \mu} \eta^{\rho \mu} (x')_\rho &= \eta_{\alpha \mu} L^\mu_{~\nu} \eta^{\sigma \nu} x_\sigma \\ \eta_{\alpha \mu} \eta^{\mu \rho} (x')_\rho &= \eta_{\alpha \mu} L^\mu_{~\nu} \eta^{\sigma \nu} x_\sigma \\ \end{aligned}}

\displaystyle{ \begin{aligned} \delta_{\alpha \rho} (x')_\rho &= \eta_{\alpha \mu} L^\mu_{~\nu} \eta^{\sigma \nu} x_\sigma \\ (x')_\alpha &= \eta_{\alpha \mu} L^\mu_{~\nu} \eta^{\sigma \nu} x_\sigma \\ \end{aligned}}

Prove that $\displaystyle{\eta_{\alpha \mu} L^\mu_{~\nu} \eta^{\sigma \nu} = \left[L^{-1}\right]^\sigma_{~\alpha}}$.

By index renaming, \displaystyle{ \begin{aligned} (x')_\mu &= \eta_{\mu \rho} L^\rho_{~\sigma} \eta^{\nu \sigma} x_\nu \\ \end{aligned}}, the question becomes

Prove that $\displaystyle{ \eta_{\mu \rho} L^\rho_{~\sigma} \eta^{\nu \sigma} = \left[L^{-1}\right]^\nu_{~\mu}}$.

Denote $\displaystyle{ \eta_{\mu \rho} L^\rho_{~\sigma} \eta^{\nu \sigma}}$ as $\displaystyle{L^{~\nu}_{\mu}}$. Then the question is simplified to

Prove that $\displaystyle{ L^{~\nu}_{\mu} = \left[L^{-1}\right]^\nu_{~\mu}}$.

\displaystyle{ \begin{aligned} (x')^\mu &= L^\mu_{~\nu} x^\nu \\ (x')_\mu &= \eta_{\mu \rho} L^\rho_{~\sigma} \eta^{\beta \sigma} x_\beta \\ \end{aligned}}

\displaystyle{ \begin{aligned} (x')^\mu (x')_\mu &= \left( L^\mu_{~\nu} x^\nu \right) \left( \eta_{\mu \rho} L^\rho_{~\sigma} \eta^{\beta \sigma} x_\beta \right) \\ (x')^\mu (x')_\mu &= \left( L^\mu_{~\nu} x^\nu \right) \left( {L^{~\beta}_{\mu}} x_\beta \right) \\ \\ \end{aligned}}

\displaystyle{ \begin{aligned} \sum_{\mu = 0}^4 (x')^\mu (x')_\mu &= \sum_{\mu = 0}^4 \sum_{\nu = 0}^4 \sum_{\beta = 0}^4 \left( L^\mu_{~\nu} x^\nu \right) \left( {L^{~\beta}_{\mu}} x_\beta \right) \\ \end{aligned}}

\displaystyle{ \begin{aligned} \sum_{\mu = 0}^4 (x')^\mu (x')_\mu &= \sum_{\mu = 0}^4 \left( \sum_{\nu = 0}^4 L^\mu_{~\nu} x^\nu \right) \left( \sum_{\beta = 0}^4 {L^{~\beta}_{\mu}} x_\beta \right) \\ \end{aligned}}

\displaystyle{ \begin{aligned} &(x')^0 (x')_0 + (x')^1 (x')_1 + (x')^2 (x')_2 + (x')^3 (x')_3 \\ &= \sum_{\mu = 0}^4 \left( L^\mu_{~0} x^0 + L^\mu_{~1} x^1 + L^\mu_{~2} x^2 + L^\mu_{~3} x^3 \right) \left( L^{~0}_{\mu} x_0 + L^{~1}_{\mu} x_1 + L^{~2}_{\mu} x_2 + L^{~3}_{\mu} x_3 \right) \\ \end{aligned}}

The right hand side has 64 terms.

Since the spacetime interval is Lorentz-invariant, $\displaystyle{ (x')^\mu (x')_\mu = x^\mu x_\mu }$. So the left hand side can be replaced by $\displaystyle{ x^\mu x_\mu }$.

\displaystyle{ \begin{aligned} &x^0 x_0 + x^1 x_1 + x^2 x_2 + x^3 x_3 \\ &= \sum_{\mu = 0}^4 \left( L^\mu_{~0} x^0 + L^\mu_{~1} x^1 + L^\mu_{~2} x^2 + L^\mu_{~3} x^3 \right) \left( L^{~0}_{\mu} x_0 + L^{~1}_{\mu} x_1 + L^{~2}_{\mu} x_2 + L^{~3}_{\mu} x_3 \right) \\ \end{aligned}}

Note that the 4 terms on the left side also appear on the right hand side.

\displaystyle{ \begin{aligned} (x')^\mu &= L^\mu_{~\nu} x^\nu \\ (x')^\mu (x')_\mu &= \left( L^\mu_{~\nu} x^\nu \right) \left( \eta_{\mu \rho} L^\rho_{~\sigma} \eta^{\beta \sigma} x_\beta \right) \\ (x')^\mu (x')_\mu &= \left( L^\mu_{~\nu} x^\nu \right) \left( {L^{~\beta}_{\mu}} x_\beta \right) \\ x^\mu x_\mu &= \left( L^\mu_{~\nu} x^\nu \right) \left( {L^{~\beta}_{\mu}} x_\beta \right) \\ \end{aligned}}

\displaystyle{ \begin{aligned} x^0 x_0 + x^1 x_1 + x^2 x_2 + x^3 x_3 &= \sum_{\mu = 0}^4 \sum_{\nu = 0}^4 \sum_{\beta = 0}^4 L^\mu_{~\nu} L^{~\beta}_{\mu} x^\nu x_\beta \\ \end{aligned}}

Since this equation is true for any coordinates, it is an identity. By comparing coefficients, we have:

1. For any terms with $\displaystyle{\nu \ne \beta}$, such as $\displaystyle{\nu = 0}$ and $\displaystyle{\beta=1}$,

\displaystyle{ \begin{aligned} \sum_{\mu = 0}^4 L^\mu_{~0} L^{~1}_{\mu} x^0 x_1 &\equiv 0 \\ \left( \sum_{\mu = 0}^4 L^\mu_{~0} L^{~1}_{\mu} \right) x^0 x_1 &\equiv 0\\ \end{aligned}}

So

\displaystyle{ \begin{aligned} \sum_{\mu = 0}^4 L^\mu_{~0} L^{~1}_{\mu} &= 0 \\ \end{aligned}}

2. For any terms with $\displaystyle{\nu = \beta}$.

\displaystyle{ \begin{aligned} x^0 x_0 + x^1 x_1 + x^2 x_2 + x^3 x_3 &\equiv \sum_{\mu = 0}^4 \sum_{\nu = 0}^4 \sum_{\beta = \nu} L^\mu_{~\nu} L^{~\beta}_{\mu} x^\nu x_\beta \\ x^0 x_0 + x^1 x_1 + x^2 x_2 + x^3 x_3 &\equiv \sum_{\mu = 0}^4 \left( L^\mu_{~0} L^{~0}_{\mu} x^0 x_0 + L^\mu_{~1} L^{~1}_{\mu} x^1 x_1 + L^\mu_{~2} L^{~2}_{\mu} x^2 x_2 + L^\mu_{~3} L^{~3}_{\mu} x^3 x_3 \right) \\ \end{aligned}}

So

\displaystyle{ \begin{aligned} \sum_{\mu = 0}^4 L^\mu_{~0} L^{~0}_{\mu} &= 1 \\ \sum_{\mu = 0}^4 L^\mu_{~1} L^{~1}_{\mu} &= 1 \\ \sum_{\mu = 0}^4 L^\mu_{~2} L^{~2}_{\mu} &= 1 \\ \sum_{\mu = 0}^4 L^\mu_{~3} L^{~3}_{\mu} &= 1 \\ \end{aligned}}

.

Denoting $\displaystyle{ \eta_{\mu \rho} L^\rho_{~\sigma} \eta^{\nu \sigma}}$ as $\displaystyle{L^{~\nu}_{\mu}}$ is misleading, because that presupposes that $\displaystyle{ \eta_{\mu \rho} L^\rho_{~\sigma} \eta^{\nu \sigma}}$ is directly related to the matrix $\displaystyle{L}$.

To avoid this bug, instead, we denote $\displaystyle{ \eta_{\mu \rho} L^\rho_{~\sigma} \eta^{\nu \sigma}}$ as $\displaystyle{M ^\nu_{~\mu}}$. So

\displaystyle{ \begin{aligned} (x')^\mu &= L^\mu_{~\nu} x^\nu \\ (x')^\mu (x')_\mu &= \left( L^\mu_{~\nu} x^\nu \right) \left( \eta_{\mu \rho} L^\rho_{~\sigma} \eta^{\beta \sigma} x_\beta \right) \\ (x')^\mu (x')_\mu &= \left( L^\mu_{~\nu} x^\nu \right) \left( M^{\beta}_{~\mu} x_\beta \right) \\ x^\mu x_\mu &= \left( L^\mu_{~\nu} x^\nu \right) \left( M^{\beta}_{~\mu} x_\beta \right) \\ \end{aligned}}

\displaystyle{ \begin{aligned} \nu \neq \mu:&~~~~~~\sum_{\mu = 0}^4 L^\mu_{~\nu} M^{\beta}_{~\mu} &= 0 \\ \nu = \mu:&~~~~~~\sum_{\mu = 0}^4 L^\mu_{~\nu} M^{\beta}_{~\mu} &= 1 \\ \end{aligned}}

— Me@2020-09-12 09:33:00 PM

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2020.09.13 Sunday (c) All rights reserved by ACHK

# Problem 2.3b2

Prove that a metric tensor is symmetric.

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Assume $\displaystyle{\eta_{\alpha\beta} \neq \eta_{\beta\alpha}}$. Because it’s irrelevant what letter we use for our indices,

$\displaystyle{\eta_{\alpha\beta}dx^{\alpha}dx^{\beta} = \eta_{\beta\alpha}dx^{\beta}dx^{\alpha}}$.

Then

$\displaystyle{\eta_{\alpha\beta}dx^{\alpha}dx^{\beta} = \frac{1}{2}(\eta_{\alpha\beta}dx^{\alpha}dx^{\beta} + \eta_{\beta\alpha}dx^{\beta}dx^{\alpha}) = \frac{1}{2} (\eta_{\alpha\beta} + \eta_{\beta\alpha})dx^{\alpha}dx^{\beta}}$

So only the symmetric part of $\displaystyle{\eta_{\alpha\beta}}$ would survive the sum. As such we may as well take $\displaystyle{\eta_{\alpha\beta}}$ to be symmetric in its definition.

— edited Jun 15 ’15 at 22:48

— rob

— answered Jun 15 ’15 at 17:52

— FenderLesPaul

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— Why is the metric tensor symmetric?

— Physics StackExchange

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1.

$\displaystyle{\eta_{\alpha\beta}dx^{\alpha}dx^{\beta} = \eta_{\beta\alpha}dx^{\beta}dx^{\alpha}}$

means that

$\displaystyle{\sum_{\alpha, \beta} \eta_{\alpha\beta}dx^{\alpha}dx^{\beta}=\sum_{\alpha, \beta}\eta_{\beta\alpha}dx^{\beta}dx^{\alpha}}$

So in

$\displaystyle{\eta_{\alpha\beta}dx^{\alpha}dx^{\beta} = \eta_{\beta\alpha}dx^{\beta}dx^{\alpha}}$,

we cannot cancel out $\displaystyle{dx^{\alpha}dx^{\beta}}$ on both sides. In other words, we do NOT assume that $\displaystyle{\eta_{\alpha\beta} = \eta_{\beta\alpha}}$ in the first place.

.

2.

$\displaystyle{\eta_{\alpha\beta}dx^{\alpha}dx^{\beta} = \frac{1}{2}(\eta_{\alpha\beta}dx^{\alpha}dx^{\beta} + \eta_{\beta\alpha}dx^{\beta}dx^{\alpha}) = \frac{1}{2} (\eta_{\alpha\beta} + \eta_{\beta\alpha})dx^{\alpha}dx^{\beta}}$

means that

$\displaystyle{\sum_{\alpha, \beta}\eta_{\alpha\beta}dx^{\alpha}dx^{\beta} = \frac{1}{2}\sum_{\alpha, \beta}(\eta_{\alpha\beta}dx^{\alpha}dx^{\beta} + \eta_{\beta\alpha}dx^{\beta}dx^{\alpha}) = \frac{1}{2} \sum_{\alpha, \beta}(\eta_{\alpha\beta} + \eta_{\beta\alpha})dx^{\alpha}dx^{\beta}}$

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3. “… only the symmetric part of $\displaystyle{\eta_{\alpha\beta}}$ would survive the sum” means that only the sum $\displaystyle{\left(\eta_{\alpha\beta} + \eta_{\beta\alpha}\right)}$ is physically meaningful.

— Me@2020-08-14 03:34:05 PM

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2020.08.14 Friday (c) All rights reserved by ACHK

# Problem 2.3b1

A First Course in String Theory

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2.3 Lorentz transformations, derivatives, and quantum operators.

(b) Show that the objects $\displaystyle{\frac{\partial}{\partial x^\mu}}$ transform under Lorentz transformations in the same way as the $\displaystyle{a_\mu}$ considered in (a) do. Thus, partial derivatives with respect to conventional upper-index coordinates $\displaystyle{x^\mu}$ behave as a four-vector with lower indices – as reflected by writing it as $\displaystyle{\partial_\mu}$.

~~~

\displaystyle{ \begin{aligned} (x')^\mu &= L^\mu_{~\nu} x^\nu \\ \frac{\partial}{\partial (x')^\mu} &= \frac{\partial x^\nu}{\partial (x')^\mu} \frac{\partial}{\partial x^\nu} \\ &= \frac{\partial x^0}{\partial (x')^\mu} \frac{\partial}{\partial x^0} + \frac{\partial x^1}{\partial (x')^\mu} \frac{\partial}{\partial x^1} + \frac{\partial x^2}{\partial (x')^\mu} \frac{\partial}{\partial x^2} + \frac{\partial x^3}{\partial (x')^\mu} \frac{\partial}{\partial x^3} \\ \end{aligned}}

The Lorentz transformation:

\displaystyle{ \begin{aligned} (x')^\mu &= L^\mu_{~\nu} x^\nu \\ \end{aligned}}

Lowering the indices to create covariant vectors:

\displaystyle{ \begin{aligned} x_\mu &= \eta_{\mu \nu} x^\nu \\ \end{aligned}}

In matrix form, covariant vectors are represented by row vectors:

\displaystyle{ \begin{aligned} \left[ x_\mu \right] &= \left( [\eta_{\mu \nu}] [x^\nu] \right)^T \\ \end{aligned}}

Change the subject:

\displaystyle{ \begin{aligned} \left[ x_\mu \right]^T &= [\eta_{\mu \nu}] [x^\nu] \\ [\eta_{\mu \nu}] [x^\nu] &= \left[ x_\mu \right]^T \\ [x^\nu] &= [\eta_{\mu \nu}]^{-1} \left[ x_\mu \right]^T \\ \end{aligned}}

With \displaystyle{ \begin{aligned} \eta^{\mu \nu} &\stackrel{\text{\tiny def}}{=} \left[ \eta_{\mu \nu} \right]^{-1} \\ \end{aligned}}, we have:

\displaystyle{ \begin{aligned} \left[ x^\nu \right] &= \left[ \eta^{\mu \nu} \right] \left[ x_\mu \right]^T \\ \end{aligned}}

\displaystyle{ \begin{aligned} x^\nu &= x_\mu \eta^{\mu \nu} \\ \end{aligned}}

Now we lower the indices in order to find the Lorentz transformation for the covariant components:

\displaystyle{ \begin{aligned} (x')^\mu &= L^\mu_{~\nu} x^\nu \\ \eta^{\rho \mu} x_\rho &= L^\mu_{~\nu} \eta^{\sigma \nu} x_\sigma \\ x_\rho &= \eta_{\rho \mu} L^\mu_{~\nu} \eta^{\sigma \nu} x_\sigma \\ \end{aligned}}

— Me@2020-07-21 10:46:32 AM

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2020.07.22 Wednesday (c) All rights reserved by ACHK

# Problem 2.3a

A First Course in String Theory

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2.2 Lorentz transformations, derivatives, and quantum operators.

(a) Give the Lorentz transformations for the components $\displaystyle{a_{\mu}}$ of a vector under a boost along the $\displaystyle{x^1}$ axis.

~~~

\displaystyle{\begin{aligned} \begin{bmatrix} c t' \\ z' \\ x' \\ y' \end{bmatrix} &= \begin{bmatrix} \gamma & -\beta \gamma & 0 & 0 \\ -\beta \gamma & \gamma & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \\ \end{bmatrix} \begin{bmatrix} c\,t \\ z \\ x \\ y \end{bmatrix} \\ \end{aligned}}

\displaystyle{ \begin{aligned} (x')^\mu &= L^\mu_{~\nu} x^\nu \\ a_\mu &= a^\nu \eta_{\mu \nu} \\ (a')_\mu &= L_\mu^{~\nu} a_\nu \\ [(a')_\mu] &= [a_\nu] [L^\mu_{~\nu}]^{-1} \\ [L^\mu_{~\nu}]^{-1} &= \begin{bmatrix} \gamma & \beta \gamma & 0 & 0 \\ \beta \gamma & \gamma & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \\ \end{bmatrix} \\ \end{aligned}}

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Another method is to start with:

\displaystyle{ \begin{aligned} a_0 &= -a^0 \\ a_1 &= a^1 \\ a_2 &= a^2 \\ a_3 &= a^3 \\ \end{aligned}}

— Me@2020-07-05 05:40:44 PM

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2020.07.05 Sunday (c) All rights reserved by ACHK

# Problem 2.2c

A First Course in String Theory

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2.2 Lorentz transformations for light-cone coordinates.

Consider coordinates $\displaystyle{x^\mu = ( x^0, x^1, x^2, x^3 )}$ and the associated light-cone coordinates $\displaystyle{x^\mu = ( x^+, x^-, x^2, x^3 )}$. Write the following Lorentz transformations in terms of the light-cone coordinates.

(c) A boost with velocity parameter $\displaystyle{\beta}$ in the $\displaystyle{x^3}$ direction.

\displaystyle{ \begin{aligned} \begin{bmatrix} c t' \\ z' \\ x' \\ y' \end{bmatrix} &= \begin{bmatrix} \gamma & -\beta \gamma & 0 & 0 \\ -\beta \gamma & \gamma & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \\ \end{bmatrix} \begin{bmatrix} c\,t \\ z \\ x \\ y \end{bmatrix} \\ \end{aligned} }

~~~

\displaystyle{ \begin{aligned} \begin{bmatrix} c t' \\ x' \\ y' \\ z' \end{bmatrix} &= \begin{bmatrix} \gamma & 0 & 0 & -\beta \gamma \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ -\beta \gamma & 0 & 0 & \gamma \\ \end{bmatrix} \begin{bmatrix} c\,t \\ x \\ y \\ z \end{bmatrix} \\ \end{aligned} }

\displaystyle{ \begin{aligned} \begin{bmatrix} 1 & 1 & 0 & 0 \\ 1 & -1 & 0 & 0 \\ 0 & 0 & \sqrt{2} & 0 \\ 0 & 0 & 0 & \sqrt{2} \\ \end{bmatrix} \begin{bmatrix} (x^+)' \\ (x^-)' \\ y' \\ z' \end{bmatrix} &= \begin{bmatrix} \gamma & 0 & 0 & -\beta \gamma \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ -\beta \gamma & 0 & 0 & \gamma \\ \end{bmatrix} \begin{bmatrix} 1 & 1 & 0 & 0 \\ 1 & -1 & 0 & 0 \\ 0 & 0 & \sqrt{2} & 0 \\ 0 & 0 & 0 & \sqrt{2} \\ \end{bmatrix} \begin{bmatrix} x^+ \\ x^- \\ y \\ z \end{bmatrix} \\ \end{aligned} }

\displaystyle{ \begin{aligned} \begin{bmatrix} (x^+)' \\ (x^-)' \\ y' \\ z' \end{bmatrix} &= \frac{1}{2} \begin{bmatrix} \gamma + 1 & \gamma - 1 & 0 & -\sqrt{2}\,\beta\,\gamma \\ \gamma - 1 & \gamma + 1 & 0 & -\sqrt{2}\,\beta\,\gamma \\ 0 & 0 & 2 & 0 \\ - \sqrt{2}\,\beta\,\gamma & - \sqrt{2}\,\beta\,\gamma & 0 & 2 \gamma \end{bmatrix} \begin{bmatrix} x^+ \\ x^- \\ y \\ z \end{bmatrix} \\ \end{aligned} }

— Me@2020-04-19 11:52:09 PM

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2020.04.21 Tuesday (c) All rights reserved by ACHK

# Problem 2.2b

A First Course in String Theory

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2.2 Lorentz transformations for light-cone coordinates.

Consider coordinates $\displaystyle{x^\mu = ( x^0, x^1, x^2, x^3 )}$ and the associated light-cone coordinates $\displaystyle{x^\mu = ( x^+, x^-, x^2, x^3 )}$. Write the following Lorentz transformations in terms of the light-cone coordinates.

(b) A rotation with angle $\displaystyle{\theta}$ in the $\displaystyle{x^1, x^2}$ plane.

\displaystyle{ \begin{aligned} \begin{bmatrix} c t' \\ x' \\ y' \\ z' \end{bmatrix} &= \begin{bmatrix} 1 & 0 & 0 & 0 \\ 0 & \cos \theta & -\sin \theta & 0 \\ 0 & \sin \theta & \cos \theta & 0 \\ 0 & 0 & 0 & 1 \\ \end{bmatrix} \begin{bmatrix} c\,t \\ x \\ y \\ z \end{bmatrix} \\ \end{aligned} }

~~~

\displaystyle{ \begin{aligned} \begin{bmatrix} \frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}} & 0 & 0 \\ \frac{1}{\sqrt{2}} & -\frac{1}{\sqrt{2}} & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \\ \end{bmatrix} \begin{bmatrix} (x^+)' \\ (x^-)' \\ y' \\ z' \end{bmatrix} &= \begin{bmatrix} 1 & 0 & 0 & 0 \\ 0 & \cos \theta & -\sin \theta & 0 \\ 0 & \sin \theta & \cos \theta & 0 \\ 0 & 0 & 0 & 1 \\ \end{bmatrix} \begin{bmatrix} \frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}} & 0 & 0 \\ \frac{1}{\sqrt{2}} & -\frac{1}{\sqrt{2}} & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \\ \end{bmatrix} \begin{bmatrix} x^+ \\ x^- \\ y \\ z \end{bmatrix} \\ \begin{bmatrix} (x^+)' \\ (x^-)' \\ y' \\ z' \end{bmatrix} &= \frac{1}{2} \begin{bmatrix} \cos\theta + 1 & 1 - \cos\theta & -\sqrt{2} \sin\theta & 0 \\ 1 - \cos\theta & \cos\theta + 1 & \sqrt{2} \sin\theta & 0 \\ \sqrt{2} \sin{\theta} & -\sqrt{2} \sin{\theta} & 2 \cos{\theta} & 0 \\ 0 & 0 & 0 & 2 \\ \end{bmatrix} \begin{bmatrix} x^+ \\ x^- \\ y \\ z \end{bmatrix} \end{aligned} }

— Me@2020-03-22 10:16:09 PM

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2020.03.23 Monday (c) All rights reserved by ACHK

# Problem 2.2a

A First Course in String Theory

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2.2 Lorentz transformations for light-cone coordinates.

Consider coordinates $\displaystyle{x^\mu = ( x^0, x^1, x^2, x^3 )}$ and the associated light-cone coordinates $\displaystyle{x^\mu = ( x^+, x^-, x^2, x^3 )}$. Write the following Lorentz transformations in terms of the light-cone coordinates.

(a) A boost with velocity parameter $\displaystyle{\beta}$ in the $\displaystyle{x^1}$ direction.

\displaystyle{ \begin{aligned} \begin{bmatrix} c t' \\ x' \\ y' \\ z' \end{bmatrix} &= \begin{bmatrix} \gamma&-\beta \gamma&0&0\\ -\beta \gamma&\gamma&0&0\\ 0&0&1&0\\ 0&0&0&1\\ \end{bmatrix} \begin{bmatrix} c\,t \\ x \\ y \\ z \end{bmatrix} \\ \end{aligned}}

~~~

\displaystyle{ \begin{aligned} \begin{bmatrix} x^+ \\ x^- \end{bmatrix} &= \frac{1}{\sqrt{2}} \begin{bmatrix} 1 & 1 \\ 1 & -1 \\ \end{bmatrix} \begin{bmatrix} x^0 \\ x^1 \end{bmatrix} \end{aligned}}

The matrix is its own inverse.

\displaystyle{ \begin{aligned} \begin{bmatrix} x^0 \\ x^1 \end{bmatrix} &= \frac{1}{\sqrt{2}} \begin{bmatrix} 1 & 1 \\ 1 & -1 \\ \end{bmatrix} \begin{bmatrix} x^+ \\ x^- \end{bmatrix} \\ \end{aligned}}

\displaystyle{ \begin{aligned} \begin{bmatrix} (x^0)' \\ (x^1)' \end{bmatrix} &= \begin{bmatrix} \frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}} \\ \frac{1}{\sqrt{2}} & -\frac{1}{\sqrt{2}} \\ \end{bmatrix} \begin{bmatrix} (x^+)' \\ (x^-)' \end{bmatrix} \\ \end{aligned}}

Apply the result to the original transformation:

\displaystyle{ \begin{aligned} \begin{bmatrix} (x^0)' \\ (x^1)' \\ (x^2)' \\ (x^3)' \end{bmatrix} &= \begin{bmatrix} \gamma&-\beta \gamma&0&0\\ -\beta \gamma&\gamma&0&0\\ 0&0&1&0\\ 0&0&0&1\\ \end{bmatrix} \begin{bmatrix} x^0 \\ x^1 \\ x^2 \\ x^3 \end{bmatrix} \\ \end{aligned}}

\displaystyle{ \begin{aligned} \begin{bmatrix} \frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}} & 0 & 0 \\ \frac{1}{\sqrt{2}} & -\frac{1}{\sqrt{2}} & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \\ \end{bmatrix} \begin{bmatrix} (x^+)' \\ (x^-)' \\ y' \\ z' \end{bmatrix} &= \begin{bmatrix} \gamma&-\beta \gamma&0&0\\ -\beta \gamma&\gamma&0&0\\ 0&0&1&0\\ 0&0&0&1\\ \end{bmatrix} \begin{bmatrix} \frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}} & 0 & 0 \\ \frac{1}{\sqrt{2}} & -\frac{1}{\sqrt{2}} & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \\ \end{bmatrix} \begin{bmatrix} x^+ \\ x^- \\ y \\ z \end{bmatrix} \end{aligned}}

\displaystyle{ \begin{aligned} \begin{bmatrix} 1 & 1 \\ 1 & -1 \\ \end{bmatrix} \begin{bmatrix} (x^+)' \\ (x^-)' \end{bmatrix} &= \begin{bmatrix} \gamma & -\beta \gamma \\ -\beta \gamma &\gamma \end{bmatrix} \begin{bmatrix} 1 & 1 \\ 1 & -1 \\ \end{bmatrix} \begin{bmatrix} x^+ \\ x^- \end{bmatrix} \end{aligned}}

\displaystyle{ \begin{aligned} \begin{bmatrix} (x^+)' \\ (x^-)' \end{bmatrix} &= \begin{bmatrix} \gamma (1-\beta) & 0 \\ 0 & \gamma (1+\beta) \\ \end{bmatrix} \begin{bmatrix} x^+ \\ x^- \end{bmatrix} \end{aligned}}

— Me@2020-02-27 07:14:19 PM

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2020.02.27 Thursday (c) All rights reserved by ACHK

# Problem 2.1b

A First Course in String Theory

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2.1 Exercises with units

(b) Explain the meaning of the unit K (degree kelvin) used for measuring temperatures, and explain its relation to the basic length, mass, and time units.

~~~

${\displaystyle {\frac {1}{T}}=\left({\frac {\partial S}{\partial U}}\right)_{V,N}}$,

where $\displaystyle{U}$ is the internal energy.
.

The units of $\displaystyle{k_B T}$ and $\displaystyle{E}$ are the same.

$\displaystyle{[k_B T] = [E]}$

In other words, the Boltzmann constant $\displaystyle{k_B}$ translates the temperature unit $\displaystyle{K}$ to the language of energy unit $\displaystyle{J}$.

However, although the temperature unit $\displaystyle{K}$ and the energy unit $\displaystyle{J}$ have the relation

$\displaystyle{k_B K = J}$,

just $\displaystyle{k_B T}$ would not give the correct value of energy $\displaystyle{E}$, not to mention that we have not yet specified of which the energy $\displaystyle{E}$ is.

.

For an ideal gas,

$\displaystyle{pV=Nk_B T}$

and the average translational kinetic energy is

${\displaystyle {\frac {1}{2}}m{\overline {v^{2}}}={\frac {3}{2}}k_BT}$

for 3 degrees of freedom. In 3D space, if there are only translational motions, there are only 3 degrees of freedom.

In other words, just the value of ${k_B T}$ itself gives no physical meaning. Instead, ${\tfrac{1}{2}k_B T}$ can be interpreted as the average translational kinetic energy of the particles in an one dimensional space. Equivalently, $\displaystyle{\tfrac{3}{2}k_BT}$ gives that in our three dimensional space.

.

Another main difference is that although energy is an extensive property, temperature is an intensive property.

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We define the temperature unit Kelvin $\displaystyle{K}$ by requiring the water triple point temperature,

$\displaystyle{T_{tp} \equiv 273.16K}$

Once this value is fixed, the Boltzmann constant $\displaystyle{k_B}$ value can be estimated by using, for example, the ideal gas law

$\displaystyle{pV = N k_B T}$,

because $\displaystyle{k_B}$ always comes with $\displaystyle{T}$.

— Me@2020-02-16 11:14:24 AM

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2020.02.16 Sunday (c) All rights reserved by ACHK

# Problem 2.1

A First Course in String Theory

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2.1 Exercises with units

Construct and evaluate a dimensionless number using the charge $\displaystyle{e}$ of the electron (as defined in Gaussian system of units), $\displaystyle{\hbar}$, and $\displaystyle{c}$. (In Heaviside-Lorentz units, the Gaussian $\displaystyle{e^2}$ is replaced by $\displaystyle{\frac{e^2}{4 \pi}}$.)

~~~

\displaystyle{ \begin{aligned} \alpha &= \frac{e^2}{(4 \pi \varepsilon_0)\hbar c} \\ \alpha &= \frac{1}{137.035\,999\,679(94)} \\ \end{aligned} }

— Me@2010

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2020.02.04 Tuesday (c) All rights reserved by ACHK

# Quick Calculation 13.2

A First Course in String Theory

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Verify that

$\displaystyle{\left[ \bar L_0^\perp, X^I (\tau, \sigma) \right] = - \frac{i}{2} \left( {\dot{X}}^I + {X^I}' \right)}$,

$\displaystyle{\left[ L_0^\perp, X^I (\tau, \sigma) \right] = - \frac{i}{2} \left( {\dot{X}}^I - {X^I}' \right)}$,

~~~

Equation (13.24):

$\displaystyle{X^{\mu} (\tau, \sigma) = x_0^\mu + \sqrt{2 \alpha'} \alpha_0^\mu \tau + i \sqrt{\frac{\alpha'}{2}} \sum_{n \ne 0} \frac{e^{-in\tau}}{n} (\alpha_n^\mu e^{i n \sigma} + \bar \alpha_n^\mu e^{-in \sigma})}$

Equation (13.39):

$\displaystyle{{\dot{X}}^- + {X^-}' = \sqrt{2 \alpha'} \sum_{n \in \mathbb{Z}} \bar \alpha_n^- e^{-in (\tau + \sigma)}}$

$\displaystyle{{\dot{X}}^- - {X^-}' = \sqrt{2 \alpha'} \sum_{n \in \mathbb{Z}} \alpha_n^- e^{-in (\tau - \sigma)}}$

Equation (13.51):

$\displaystyle{\left[\bar L_m^\perp, \bar \alpha_n^J \right] = - n \bar{\alpha}_{m+n}^J}$

$\displaystyle{\left[L_m^\perp, \alpha_n^J \right] = - n \alpha_{m+n}^J}$

Equation (13.52):

$\displaystyle{\left[L_m^\perp, \bar \alpha_n^J \right] = 0}$

$\displaystyle{\left[\bar L_m^\perp, \alpha_n^J \right] = 0}$

Equation (13.53):

$\displaystyle{\left[ \bar L_m^\perp, x_0^I \right] = - i \sqrt{\frac{\alpha'}{2}} \bar \alpha^I_m}$

$\displaystyle{\left[ L_m^\perp, x_0^I \right] = - i \sqrt{\frac{\alpha'}{2}} \alpha^I_m}$

.

$\displaystyle{\left[ \bar L_0^\perp, X^I (\tau, \sigma) \right]}$,

$\displaystyle{= \left[ \bar L_0^\perp, x_0^I + \sqrt{2 \alpha'} \alpha_0^I \tau + i \sqrt{\frac{\alpha'}{2}} \sum_{n \ne 0} \frac{e^{-in\tau}}{n} (\alpha_n^I e^{i n \sigma} + \bar \alpha_n^I e^{-in \sigma}) \right]}$

$\displaystyle{= - i \sqrt{\frac{\alpha'}{2}} \bar \alpha^I_0 + i \sqrt{\frac{\alpha'}{2}} \sum_{n \ne 0} \frac{e^{-in\tau}}{n} e^{-in \sigma} (-n \bar \alpha_n^I)}$

$\displaystyle{= - i \sqrt{\frac{\alpha'}{2}} \bar \alpha^I_0 - i \sqrt{\frac{\alpha'}{2}} \sum_{n \ne 0} e^{-in(\tau + \sigma)} ( \bar \alpha_n^I)}$

$\displaystyle{= - \frac{i}{2} \sqrt{2\alpha'} \sum_{n \in \mathbb Z} \bar \alpha_n^I e^{-in(\tau + \sigma)}}$

— Me@2020-01-06 11:30:38 PM

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2020.01.06 Monday (c) All rights reserved by ACHK

# Quick Calculation 13.1

A First Course in String Theory

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Verify that

$\displaystyle{\left[ \bar L_m^\perp, x_0^I \right] = - i \sqrt{\frac{\alpha'}{2}} \bar \alpha^I_m}$,

$\displaystyle{\left[ L_m^\perp, x_0^I \right] = - i \sqrt{\frac{\alpha'}{2}} \alpha^I_m}$.

~~~

Equation (13.37):

$\displaystyle{\bar L_m^\perp = \frac{1}{2} \sum_{p \in \mathbf{Z}} \bar \alpha_p^J \bar \alpha_{n-p}^J}$,$\displaystyle{~~~L_m^\perp = \frac{1}{2} \sum_{p \in \mathbf{Z}} \alpha_p^J \alpha_{n-p}^J}$.

.

\displaystyle{ \begin{aligned} \left[ \bar L_m^\perp, x_0^I \right] &= \frac{1}{2} \sum_{p \in \mathbb{Z}} \left[ \bar \alpha^J_p \bar \alpha^J_{m-p}, x_0^I \right] \\ &= \frac{1}{2} \sum_{p \in \mathbb{Z}} \bar \alpha^I_p \left[ \bar \alpha^J_{m-p}, x_0^I \right] + \frac{1}{2} \sum_{p \in \mathbb{Z}} \left[ \bar \alpha^J_p, x_0^I \right] \bar \alpha^I_{m-p} \\ &= \frac{1}{2} \bar \alpha^J_m \left[ \bar \alpha^J_{0}, x_0^I \right] + \frac{1}{2} \left[ \bar \alpha^J_0, x_0^I \right] \bar \alpha^J_{m} \\ \end{aligned} }

.

By Equation (13.33):

\displaystyle{ \begin{aligned} \left[ \bar L_m^\perp, x_0^I \right] &= - \frac{1}{2} \bar \alpha^J_m \left[ i \sqrt{\frac{\alpha'}{2}} \eta^{IJ} \right] - \frac{1}{2} \left[ i \sqrt{\frac{\alpha'}{2}} \eta^{IJ} \right] \bar \alpha^J_{m} \\ &= - i \sqrt{\frac{\alpha'}{2}} \eta^{IJ} \bar \alpha^I_m \\ \end{aligned} }

.

Since $I$ and $J$ are transverse coordinate indices, neither of them can be zero.

\displaystyle{ \begin{aligned} \left[ \bar L_m^\perp, x_0^I \right] &= - i \sqrt{\frac{\alpha'}{2}} \bar \alpha^I_m \\ \end{aligned} }

— Me@2019-12-25 10:56:15 AM

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2019.12.25 Wednesday (c) All rights reserved by ACHK

# Problem 13.5b

A First Course in String Theory

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13.6 Unoriented closed strings

This problem is the closed string version of Problem 12.12. The closed string $\displaystyle{X^{\mu} (\tau, \sigma)}$ with $\displaystyle{\sigma \in [0, 2 \pi]}$ and fixed $\displaystyle{\tau}$ is a parameterized closed curve in spacetime. The orientation of a string is the direction of the increasing $\displaystyle{\sigma}$ on this curve.

Introduce an orientation reversing twist operator $\displaystyle{\Omega}$ such that

$\displaystyle{\Omega X^I(\tau, \sigma) \Omega^{-1}} = X^I (\tau, 2 \pi - \sigma)$

Moreover, declare that

$\displaystyle{\Omega x_0^- \Omega^{-1} = x_0^-}$

$\displaystyle{\Omega p^+ \Omega^{-1} = p^+}$

(b) Used the closed string oscillator expansion (13.24) to calculate

$\displaystyle{\Omega x_0^I \Omega^{-1}}$

$\displaystyle{\Omega \alpha_0^I \Omega^{-1}}$

$\displaystyle{\Omega \alpha_n^I \Omega^{-1}}$

$\displaystyle{\Omega \bar \alpha_n^I \Omega^{-1}}$

~~~

Equation (13.24):

$\displaystyle{X^{\mu} (\tau, \sigma) = x_0^\mu + \sqrt{2 \alpha'} \alpha_0^\mu \tau + i \sqrt{\frac{\alpha'}{2}} \sum_{n \ne 0} \frac{e^{-in\tau}}{n} (\alpha_n^\mu e^{i n \sigma} + \bar \alpha_n^\mu e^{-in \sigma})}$

.

\displaystyle{\begin{aligned} X^{\mu} (\tau, \sigma) &= x_0^\mu + \sqrt{2 \alpha'} \alpha_0^\mu \tau + i \sqrt{\frac{\alpha'}{2}} \sum_{n \ne 0} \frac{e^{-in\tau}}{n} (\alpha_n^\mu e^{i n \sigma} + \bar \alpha_n^\mu e^{-in \sigma}) \\ X^I (\tau, 2 \pi - \sigma) &= x_0^I + \sqrt{2 \alpha'} \alpha_0^I \tau + i \sqrt{\frac{\alpha'}{2}} \sum_{n \ne 0} \frac{e^{-in\tau}}{n} \left( \alpha_n^I e^{- in\sigma} + \bar \alpha_n^I e^{i n \sigma)} \right) \\ \end{aligned}}

.

$\displaystyle{\Omega X^I(\tau, \sigma) \Omega^{-1}} = X^I (\tau, 2 \pi - \sigma)$

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By comparing $\displaystyle{\Omega X^I(\tau, \sigma) \Omega^{-1}}$ with $\displaystyle{X^I (\tau, 2 \pi - \sigma)}$, we have:

\displaystyle{\begin{aligned} \Omega x_0^I \Omega^{-1} &= x_0^I \\ \Omega \alpha_0^I \Omega^{-1} &= \alpha_0^I \\ \Omega \alpha_n^I \Omega^{-1} &= \bar \alpha_n^I \\ \Omega \bar \alpha_n^I \Omega^{-1} &= \alpha_n^I \\ \end{aligned}}

— Me@2019-11-24 04:33:23 PM

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2019.11.24 Sunday (c) All rights reserved by ACHK

# Problem 13.6b

A First Course in String Theory | Topology, 2 | Manifold, 2

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13.6 Orientifold Op-planes

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In the mathematical disciplines of topology, geometry, and geometric group theory, an orbifold (for “orbit-manifold”) is a generalization of a manifold. It is a topological space (called the underlying space) with an orbifold structure.

The underlying space locally looks like the quotient space of a Euclidean space under the linear action of a finite group.

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In string theory, the word “orbifold” has a slightly new meaning. For mathematicians, an orbifold is a generalization of the notion of manifold that allows the presence of the points whose neighborhood is diffeomorphic to a quotient of $\displaystyle{\mathbf{R}^n}$ by a finite group, i.e. $\displaystyle{\mathbf{R}^n/\Gamma}$. In physics, the notion of an orbifold usually describes an object that can be globally written as an orbit space $\displaystyle{M/G}$ where $\displaystyle{M}$ is a manifold (or a theory), and $\displaystyle{G}$ is a group of its isometries (or symmetries) — not necessarily all of them. In string theory, these symmetries do not have to have a geometric interpretation.

— Wikipedia on Orbifold

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In mathematics, a manifold is a topological space that locally resembles Euclidean space near each point. More precisely, each point of an $\displaystyle{n}$-dimensional manifold has a neighborhood that is homeomorphic to the Euclidean space of dimension $\displaystyle{n}$.

— Wikipedia on Manifold

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In topology and related branches of mathematics, a topological space may be defined as a set of points, along with a set of neighbourhoods for each point, satisfying a set of axioms relating points and neighbourhoods. The definition of a topological space relies only upon set theory and is the most general notion of a mathematical space that allows for the definition of concepts such as continuity, connectedness, and convergence. Other spaces, such as manifolds and metric spaces, are specializations of topological spaces with extra structures or constraints.

— Wikipedia on Topological space

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2019.09.26 Thursday ACHK

# Problem 13.6

A First Course in String Theory

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13.6 Orientifold Op-planes

(a) For an O23-plane the two normal directions $\displaystyle{x^{24}, x^{25}}$ can be represented by a plane. A closed string at a fixed $\tau$ appears as a parameterized closed curve $\displaystyle{X^a(\tau, \sigma)}$ in this plane. Draw such an oriented closed string that lies fully in the first quadrant of the $\displaystyle{(x^{24}, x^{25})}$ plane. Draw also the string $\displaystyle{\tilde{X}^a(\tau, \sigma) = -X^a(\tau, 2\pi - \sigma)}$.

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This one is wrong.

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— Me@2019-08-26 10:31:07 PM

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2019.08.26 Monday (c) All rights reserved by ACHK

# Quick Calculation 15.1.2

A First Course in String Theory

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Recall that a group is a set which is closed under an associative multiplication; it contains an identity element, and each element has a multiplicative inverse. Verify that $\displaystyle{U(1)}$ and $\displaystyle{U(N)}$, as described above, are groups.

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# Definition

A group is a set, G, together with an operation $\displaystyle{\bullet}$ (called the group law of G) that combines any two elements a and b to form another element, denoted $\displaystyle{a \bullet b}$ or $\displaystyle{ab}$. To qualify as a group, the set and operation, $\displaystyle{(G, \bullet)}$, must satisfy four requirements known as the group axioms:

Closure

For all a, b in G, the result of the operation, $\displaystyle{a \bullet b}$, is also in G.

Associativity

For all a, b and c in G, $\displaystyle{(a \bullet b) \bullet c = a \bullet (b \bullet c)}$.

Identity element

There exists an element e in G such that, for every element a in G, the equation $\displaystyle{e \bullet a = a \bullet e = a}$ holds. Such an element is unique, and thus one speaks of the identity element.

Inverse element

For each a in G, there exists an element b in G, commonly denoted $\displaystyle{a^{-1}}$ (or $\displaystyle{-a}$, if the operation is denoted “+”), such that $\displaystyle{a \bullet b = b \bullet a = e}$, where e is the identity element.

— Wikipedia on Group (mathematics)

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The axioms for a group are short and natural… Yet somehow hidden behind these axioms is the monster simple group, a huge and extraordinary mathematical object, which appears to rely on numerous bizarre coincidences to exist. The axioms for groups give no obvious hint that anything like this exists.

— Richard Borcherds in Mathematicians: An Outer View of the Inner World

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2019.07.28 Sunday ACHK