Heat kernel property We define the heat propagator on a Riemannian manifold $M$: $$e^{-t \Delta_g}: L^2(M) \rightarrow L^2(M)$$ $$e^{-t \Delta_g} f(x) = \int_M p(x,y,t) f(y) \,dV(y)$$ where $p(x,y,t)$ is the fundamental solution to the heat equation on $M$. I want to prove that $e^{-t \Delta_g} \circ e^{-s \Delta_g} = e^{-(t+s) \Delta_g}$. What I have so far: Fix $f \in L^2(M)$ $$e^{-t \Delta_g} \circ e^{-s \Delta_g} (f(x)) = e^{-t \Delta_g} \left(\int_M p(x,y,t) f(y) \, dV(y)\right)$$ $$= \int_M p(y,z,s) \left(\int_M p(x,y,t) f(y) \, dV(y) \right) \, dV(z)$$ I know that $\int_M p(x,y,t)p(y,z,s)\, dV(y) = p(x,z,t+s)$ but I'm not sure how to use it.

Actually you should have

$$e^{-t \Delta_g} \circ e^{-s\Delta_g} f(x) = \int_M p(x,y,t) (e^{-s\Delta_g} f)(y) \,dV(y)$$

$$ = \int_M p(x, y, t)\bigg( \int_M p(y, z, s) f(z) dV(z) \bigg)dV(y)$$

$$ = \int_M \bigg( \int_M p(x, y, t) p(y, z, s)dV(y)\bigg) f(z) dV(z) .$$

From here I think you will know how to proceed.