Graviton Degrees of Freedom

A symmetric square matrix of the given dimension has $\frac12 (d-2) (d-1)$ degrees of freedom. To see this, one can use the formula for triangular numbers. Since the matrix is symmetric, the entries below the diagonal are not independent. The entries up to the diagonal are a triangular number (literally, since they are in the shape of a triangle) equal to $1+2+\ldots + d-2 = \frac12 (d-2)(d-1)$ . For the graviton, the traceless condition removes exactly one of these degrees of freedom (since it fixes one diagonal entry to cancel the sum of the others). Thus we have $\frac12 (d-2)(d-1) - 1 = \frac12 d(d-3)$ degrees of freedom.

See also string theory wiki . From the string-theoretic approach, one gets this square matrix from the $d-2$ degrees of freedom of the oscillators $\alpha^I_{-1}$ in light-cone coordinates (since the first two of $d$ coordinates are fully fixed by the theory).

From the approach of gravitational waves, it seems like the relevant matrix is the traceless, symmetric part of the tensor perturbations $s_{ij}$ , which is $d-1 \times d-1$ . This is the relevant matrix, but additional components are fixed by the requirement that the gravitons be massless. From a group theoretic approach, one says that a massless spin-2 particle transforms in the traceless symmetric tensor representation of $SO(D-2)$ , which has the desired number of degrees of freedom.