The <bosonic string mass spectrum> is
$$
M^2=\frac4{\alpha'}(N-a)
=\frac4{\alpha'}(\widetilde N-a),
\qquad N=\widetilde N.
$$
With Lorentz invariance, $a=1$ and $D=26$, so $N=0$ is tachyonic, $N=1$ is massless, and higher levels are massive. To display the degeneracies in general notation, put $d=D-2$. The numbers of states in one chiral sector at levels $N=0,1,2,3$ are
$$
d_0=1,\qquad
d_1=d,\qquad
d_2=\frac{d(d+3)}2,\qquad
d_3=\frac{d(d+1)(d+8)}6.
$$
Level matching pairs any left state with any right state at the same level, so the total closed-string degeneracies are
$$
1,\qquad
d^2,\qquad
\left[\frac{d(d+3)}2\right]^2,\qquad
\left[\frac{d(d+1)(d+8)}6\right]^2.
$$
At $D=26$ these are $1$, $576$, $104976$, and $10240000$. If “first four levels” is taken to exclude the ground state, the next chiral degeneracy is
$$
d_4=d+d^2+\frac{d(d+1)}2+\frac{d^2(d+1)}2+\binom{d+3}{4},
$$
which is $25650$ at $d=24$, giving total degeneracy $25650^2=657922500$.

Solved by gpt-5.6-sol high.

Solution

Past exam of the mathematics course of the University of Cambridge / 2026 / iii / Paper 306 / 2 / c / Solution

For a closed bosonic string,
$$
M^2=\frac4{\alpha'}(N-a)=\frac4{\alpha'}(\widetilde N-a),
\qquad N=\widetilde N.
$$


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