Consider a finite complex $E$ of (holomorphic) vector bundles on a (complex) manifold $X$, i.e, the complex is of the form $$ 0 \to E_N \to E_{N-1} \to \dots \to {E_0} \to 0, $$ where the bundles are equipped with connections $D_i$. By K-theory, one may consider the complex $E$ as the alternating sum $\sum_i (-1)^i [E_i]$, and it is then natural to define the Chern character (as a form) as $ch(E,D) := \sum_{i=0}^N (-1)^i ch(E_i,D_i)$, and the Chern form as $c(E,D) := \prod_{i=0}^N c(E_i,D_i)^{(-1)^i}$, where $ch(E_i,D)$ and $c(E_i,D)$ denote the Chern character and Chern form of $(E_i,D_i)$.

Alternatively, for a fixed $k$, one may express the Chern character as a polynomial in the Chern forms, $ch_k = S_k(c_1,\dots,c_k)/k!$, where $S_k$ is the polynomial which expresses the Newton polynomials in terms of the elementary symmetric polynomials, i.e., what is sometimes called the Hirzebruch-Newton polynomial. For example, $S_1(t_1)=t_1$, $S_2(t_1,t_2)=t_1^2-2t_2$ etc. With the help of the polynomials $S_k$ above, one could alternatively define a Chern character of $E$ by $\widetilde{ch}_k(E,D)=S_k(c_1(E,D),\dots,c_k(E,D))/k!$.

I have only found mentioned in passing or implicitly that these definitions coincide, i.e., $ch_k(E,D) = \widetilde{ch}_k(E,D)$, but not any precise argument. Does anyone know of a convenient reference or proof of this fact?

  • $\begingroup$ Try the refs in mathoverflow.net/questions/345437/… $\endgroup$ – Tom Copeland Apr 14 at 16:32
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    $\begingroup$ That question is one of the pages where I have ended up in my search for a reference, but unfortunately it didn't lead me to an answer. $\endgroup$ – Richard L Apr 14 at 16:54

I have not yet found a reference for this fact at the level of forms, but at least I think I found a rather simple argument for this fact.

By the generating series related to Newton's identities, if $e_k(x_1,\dots,x_n)$ denote the elementary symmetric polynomial of degree $k$ and $p_k(x_1,\dots,x_n)=x_1^k+\dots+x_n^k$ denote the power sum polynomial, then $$\ln(1+e_1t+e_2t^2+\dots) = \sum_{k=1}^\infty (-1)^{k-1} \frac{p_k t^k}{k} = \sum_{k=1}^\infty (-1)^{k-1} \frac{S_k(e_1,\dots,e_k) t^k}{k}$$ (where the first equality can be proven by a calculation starting with $(d/dt)\ln(\prod(1+x_it))$). Since the $e_i$ are algebraically independent, the same identity holds when the $e_i$ are replaced by any elements in some commutative ring, like the ring of differential forms of even degree.

In particular, it follows that if $()_k$ denotes the part of a form of degree $2k$, then $$ch_k(E,D) = \frac{(-1)^{k-1}}{(k-1)!}\ln\left( \prod_i (c(E_i) ,D_i)^{(-1)^i} \right)_k = \frac{(-1)^{k-1}}{(k-1)!} \sum_i (-1)^i (\ln c(E_i,D_i))_k = \sum_i (-1)^i ch_k(E_i,D_i)_k = \widetilde{ch}_k(E,D).$$


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