Is there a geometric picture between Chern class and Chern character?

Question

I'm new to the characteristics classes. When I learn the definition of Chern class and Chern character, the total Chern class is defined as: $$c(\mathcal F)=\det(I+\frac{i \mathcal F}{2\pi}),$$ where $\mathcal F$ is the curvature matrix or field strength, $i=\sqrt{-1}$. The total Chern character is defined as: $$\operatorname{ch}(\mathcal F)=\operatorname{tr}(\exp(\frac{i \mathcal F}{2\pi})).$$ You see there is a $\det$ and a trace, which remind me of the famous identity in Lie group: $\det(\exp(A))=\exp(\operatorname{tr}(A))$, which leads to the fact that there is a $\det(A)=1$ in definition of $\mathrm{SL}$, so there is a trace=0 in $\mathrm{sl}$.

So my question is, is there also a similar geometric picture between Chern class and Chern character? similar to the relationship between a Lie group and its Lie algebra? I see only their algebraic equivalence in the textbooks I can find. Thank you.

Answer 1

I do not think that this is the correct 'compatibility', but both invariants certainly follow a similar geometric construction, which lives in the realm of supergeometry. The total Chern class can be build as $\det\circ \exp_G$ where $\exp_G:\mathfrak{g}\to G$ is the exponential map of a (infinite dimensional super) Lie-group. The Chern character might be interpreted as the (super) trace of a (super) parallel transport. As far as I know the last statement is due to Fei Han.

The following is a naive and probably incorrect description of these ideas and is avoiding the language of supermanifolds.

Let's setup notation first: Let $\pi:E \to M$ be a (smooth) complex vector bundle over a manifold with given connection $\nabla$. From this one can build a superconnection (a differential operator): $$d^\nabla:\Omega(M,E) \to \Omega(M,E).$$ Note that $\Omega(M,E)$ is a supermodule over the superalgebra $\Omega(M)$ and $d^\nabla$ is an odd operator. Now $(d^\nabla)^2 = \frac{1}{2}[d^\nabla,d^\nabla]$ is $\Omega(M)$-linear (a differential operator of order $0$) and is given by multiplication with the curvature $\mathcal{F} \in \Omega^2(M,\mathrm{End}(E))$. Moreover one has the Bianchi identity: $(d^\nabla)^3 = 0$.

Now let $G$ be the group of $\Omega(M)$-linear automorphisms of $\Omega(M,E)$. Set $A := \mathcal{F}/(2\pi i) $ and define two paths in $G$: \begin{align*}\alpha(t) &:= 1 + t A\\ \beta(t) &:= \exp(tA)\end{align*} $\alpha$ is the essentially the one parameter subgroup of $G$ which is tangent to $A$ at the identity and therefore we might set $\exp_G(A) = \alpha(1) = 1+\mathcal{F}/(2\pi i)$. Note that there is a determinant $\det:G \to \Omega(M)^\times$ and we have $c(E,\nabla)=\det(\exp_G(A))$.

$\beta$ essentially satisfies a differential equation belonging to a parallel transport (not clear here what the loop is). That is we have for $v \in \Omega(M,E)$ that $ d^\nabla \exp(tA)v = td^\nabla A \cdot\exp(tA)v = 0$ by the Bianchi identity $d^\nabla A = 0$. Here the parallel transport is a $\Omega(M)$-linear automorphism $P: \Omega(M,E) \to \Omega(M,E)$ and the Chern character is given by its trace.

Remarks:

1. You do have $\det(\exp_G(A)) = \exp_{\Omega(M)^\times}(\mathrm{tr}_\mathfrak{g}(A))$ but this does not connect to the Chern character.

2. I am using a different sign convention for the total Chern class ($i^{-1} = -i$). I strongly believe one should not seperate $2\pi$ from $i$ here.