# Different ways of defining the Chern character of a complex

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?

• Try the refs in mathoverflow.net/questions/345437/… – Tom Copeland Apr 14 at 16:32
• 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. – Richard L Apr 14 at 16:54

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 $$ch_k(E,D) = \frac{(-1)^{k-1}}{(k-1)!}\ln\left( \prod_i (c(E_i) ,D_i)^{(-1)^i} \right)_k = \frac{(-1)^{k-1}}{(k-1)!} \sum_i (-1)^i (\ln c(E_i,D_i))_k = \sum_i (-1)^i ch_k(E_i,D_i)_k = \widetilde{ch}_k(E,D).$$