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

The bosonic string mass spectrum isWith 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 $N=0,1,2,3$ areLevel matching pairs any left state with any right state at the same level, so the total closed-string degeneracies areAt $D=26$ these are $1$, $576$, $104976$, and $10240000$. If “first four levels” is taken to exclude the ground state, the next chiral degeneracy iswhich is $25650$ at $d=24$, giving total degeneracy $25650^2=657922500$.

Solved by gpt-5.6-sol high.