# Problem 14.5d3

Counting states in heterotic SO(32) string theory | A First Course in String Theory

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d) Write a generating function $\displaystyle{f_L(x) = \sum_{r} a(r) x^r}$ for the full set of GSO-truncated states in the left-moving sector (include both NS’+ and R’+ states).

Use the convention where $\displaystyle{a(r)}$ counts the number of states with $\displaystyle{\alpha' M_L^2 = r}$.

Use $\displaystyle{f_L(x)}$ and an algebraic manipulator to find the total number of states in heterotic string theory at $\displaystyle{\alpha' M_L^2 = 8}$.

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— This answer is my guess. —

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\displaystyle{ \begin{aligned} \alpha' M_L^2: \end{aligned}}

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\displaystyle{ \begin{aligned} &f_{L, NS'+}(x) \\ &= a_{NS'+} (r) x^r \\ &= \frac{1}{2x} \left[ \prod_{r=1}^\infty \frac{(1 + x^{r-\frac{1}{2}})^{32}}{(1 - x^r)^8} + \prod_{r=1}^\infty \frac{(1 - x^{r-\frac{1}{2}})^{32}}{(1 - x^r)^8} \right] \\ & \\ &= \frac{1}{x} + 504 + 40996 x + 1384320 x^{2} + ... \\ \end{aligned}}

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\displaystyle{\begin{aligned} &f_{L, R'+}(x) \\ &= a_{R'+} (r) x^r \\ &= 2^{15} x \prod_{r=1}^\infty \frac{(1 + x^{r})^{32}}{(1 - x^r)^8} \\ & \\ &= 32768 \, x+1310720 \, x^{2}+27131904 \, x^{3}+387973120 \, x^{4}+4312727552 \, x^{5} + ... \end{aligned}}

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\displaystyle{ \begin{aligned} \alpha' M_R^2: \end{aligned}}

\displaystyle{ \begin{aligned} f_{NS+}(x) &= 8 + 128 \, x + 1152 \, x^{2} + 7680 \, x^{3} + 42112 \, x^{4} + ... \\ \end{aligned}}

\displaystyle{ \begin{aligned} f_{R-}(x) &= 8 + 128 x + 1152 x^{2} + 7680 x^{3} + 42112 x^{4} + ... \\ \end{aligned}}

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So the total number of states in heterotic string theory at \displaystyle{ \begin{aligned} \alpha' M^2 = 8 \end{aligned}} is

\displaystyle{ \begin{aligned} &\left(1384320 + 1310720 \right) \times \left(1152 + 1152\right) \\ \end{aligned}}.

\displaystyle{ \begin{aligned} &= 6209372160 \\ \end{aligned}}.

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— This answer is my guess. —

— Me@2019-01-26 04:49:37 PM

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