This question is mainly about Section 5.2 of the book "Heat Kernels and Dirac Operators" by Berline, Getzler and Vergne.

Let $M$ be a compact Riemannian manifold without boundary and $P\rightarrow M$ be a principal bundle with compact structure group $G$. Let $E$ be a finite dimensional complex vector space and $\rho: G\rightarrow End(E)$ be a representation, then we get an associated vector bundle $\mathcal{E}:= P\times_G E$ over $M$.

If we have a connection on the principal bundle $P$, then we have the associated connection on the associated bundle $\mathcal{E}$, then we have the Laplacian operator $\Delta^{\mathcal{E}}$, which acts on $\Gamma(M,\mathcal{E})\cong (C^{\infty}(P)\otimes E)^G$.

On the other hand, the connection on $P$, together with the Riemannian metric on $M$ and the invariant metric on $G$, gives a Riemannian metric on the bundle (considered as a manifold) $P$. therefore we have the scalar Laplacian $\Delta^P$ acting on $C^{\infty}(P)$. Furthermore $\Delta^P\otimes 1$ acts on $C^{\infty}(P)\otimes E$.

By some computation we get the Proposition 5.6 of that book: The Laplacian $\Delta^{\mathcal{E}}$ coincide with $\Delta^P\otimes 1+1\otimes \text{Cas}$ on $(C^{\infty}(P)\otimes E)^G$, where Cas is the Casimir operator on $E$.

Then we have the heat kernel $e^{-t\Delta^{\mathcal{E}}}\in \Gamma (M\times M, \mathcal{E}\boxtimes \mathcal{E}^*)$ and we can pull it back to $P$ and get a function $e^{-t\Delta^{\mathcal{E}}}\in C^{\infty}(P\times P)\otimes End(E)$.

On the other hand the scalar Laplacian $\Delta^P$ also has a heat kernel $ e^{-t\Delta^P}\in C^{\infty}(P\times P)$. The book claims that by the above Proposition 5.6 we have $$ e^{-t\Delta^{\mathcal{E}}}= e^{-t\text{Cas}}e^{-t\Delta^P}. $$

My first question is: in what sense the above equality holds? Since $\Delta^{\mathcal{E}}$ is defined only on $(C^{\infty}(P)\otimes E)^G$ and $\Delta^P$ is defined on the whole $C^{\infty}(P)\otimes E$, it's not obvious that the two heat kernels are equal as elements in $C^{\infty}(P\times P)\otimes End(E)$. My guess is that they are equal as operator semigroups acting on $(C^{\infty}(P)\otimes E)^G$.

Then comes Proposition 5.7: If $p_1$ and $p_2$ are two points on $P$, then $$ < p_1|e^{-t\Delta^{\mathcal{E}}}|p_2 >= e^{-t\text{Cas}} \int_G < p_1|e^{-t\Delta^P}|p_2 g >\rho(g)^{-1}dg. $$

The proof of Proposition 5.7 in the book is to pick a test function $\phi\in (C^{\infty}(P)\otimes E)^G$ and we know $$ e^{-t\Delta^{\mathcal{E}}}\phi=e^{-t\text{Cas}}e^{-t\Delta^P}\phi $$ Then we calculate the right hand side and get the result.

My second question is: Is it sufficient to prove the proposition using only $\phi\in (C^{\infty}(P)\otimes E)^G$? It seems that the $G$ invariant function is not enough. By a similar argument we can "prove" that $$ < p_1|e^{-t\Delta^{\mathcal{E}}}|p_2 >= e^{-t\text{Cas}}< p_1|e^{-t\Delta^P}|p_2 > $$ but this is not true in general. Then what have I ignored in this proof of Proposition 5.7?


2 Answers 2


Your understanding is correct. All the identify is only for the $G$-invariant section.


About first question:

$e^{-t\Delta^\mathcal{E}} = e^{-t\mathrm{Cas}}e^{-t\Delta^P}$ is a equation between elements of $\mathrm{End}(C^\infty(P,E)^G)$. More precisely, $\alpha e^{-t\Delta^\mathcal{E}}\alpha^{-1} = e^{-t\mathrm{Cas}}e^{-t\Delta^P}$, where $\alpha: \Gamma(M,\mathcal{E}) \xrightarrow{\simeq} C^\infty(P,E)^G$.

About second question:

$\langle p|e^{-t\Delta^\mathcal{E}}|q\rangle = e^{-t\mathrm{Cas}}\int_G\langle p|e^{-t\Delta^P}|qg\rangle\rho(g)^{-1}|dg|$ holds for all $p,q \in P$.

First, $\Gamma(M\times M,\mathcal{E}\boxtimes\mathcal{E}^*)) \simeq C^\infty(P\times P,\mathrm{End}(E))^{G\times G}$, where $G\times G$ acts on $\mathrm{End}(E)$ by $G\times G\times\mathrm{End}(E) \ni (g,h,A) \mapsto \rho(g)A\rho(h)^{-1} \in \mathrm{End}(E)$. $\langle p|e^{-t\Delta^\mathcal{E}}|q\rangle \in C^\infty(P\times P,\mathrm{End}(E))^{G\times G}$ is defined as the correspondent of the heat kernel $\langle x|e^{-t\Delta^\mathcal{E}}|y\rangle \in \Gamma(M\times M,\mathcal{E}\boxtimes\mathcal{E}^*))$. $\langle p|e^{-t\Delta^\mathcal{E}}|q\rangle$ is not the "kernel" of the operator $\alpha e^{-t\Delta^\mathcal{E}}\alpha^{-1}$, but $|G|$ times of the kernel of the operator $\alpha e^{-t\Delta^\mathcal{E}}\alpha^{-1}P_G$. Here, $P_G: C^\infty(P,E) \to C^\infty(P,E)^G$ is defined by $(P_G s)(p) = \frac{1}{|G|}\int_G\rho(g)s(pg)|dg|$.

Let's see $|G|$ times of the kernels of the equation $e^{-t\Delta^\mathcal{E}}P_G = e^{-t\mathrm{Cas}}e^{-t\Delta^P}P_G$. $|G|$ times of the kernel of the left hand side is $\langle p|e^{-t\Delta^\mathcal{E}}|q\rangle$, and that of the right hand side is $e^{-t\mathrm{Cas}}\int_G\langle p|e^{-t\Delta^P}|qg^{-1}\rangle\rho(g)|dg| = e^{-t\mathrm{Cas}}\int_G\langle p|e^{-t\Delta^P}|qg\rangle\rho(g)^{-1}|dg|$.


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