Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Let $B$ be the graded ring $\bigoplus_i B^i$ (with $B^k B^l \subset B^{k+l}$), and $B_f$ the multiplicative group of all formal sums $1 + b_1 + b_2 + \cdots$ where $b_i \in B^i$ for all $i$.

The idea when talking about genera (such as the Todd genus or the $L$ genus) is that we can take $B$ to be $H^{2 \bullet}(X,\mathbb{Q})$ for example, and then a typical element of $B_f$ is something like the total Chern class.

Now a genus corresponds to a multiplicative sequence $(K_n)$, where each $K_n$ is a polynomial in n variables over $B$ that is homogeneous (with respect to the grading), so that $K_n(t x_1, t^2 x_2, \ldots, t^n x_n) = t^n K_n(x_1, \ldots, x_n)$. This corresponds to the idea that for an $n$ dimensional complex manifold, the evaluation of $K_n$ on the fundamental class should be some number, so that we want this homogeneity to ensure that $K_n$ corresponds to some element of the top cohomology group.

Given this, we can form an element $K(b)$ in $B_f$ for each element $b = 1 + b_1 + b_2 + \cdots$ of $B_f$ by $K(b) = 1 + K_1(b_1) + K_2(b_1,b_2) + \cdots$.
Then multiplicativity of the sequence $(K_n)$ means that $K(bc) = K(b)K(c)$, which is what you would want in algebraic topology to get multiplicativity of the numbers you get, from multiplicativity of the total Chern classes.

Now, all the books I have been reading about this say that there is an essentially unique way to associate such a multiplicative sequence to a formal power series, and that all multiplicative sequences come from this. But I don't understand precisely how; I don't even know over what ring the formal power series should be defined and even less how one can find a multiplicative sequence out of that.

It's easy enough when $B = H^{2 \bullet}(X,\mathbb{Q})$ and you're working with Chern classes, as the Chern classes of a direct sum of lines bundles are given by the symmetric polynomials in the first Chern classes. Then, given a formal power series $Q(x) \in \mathbb{Q}[[x]]$, one can form the product $\prod_i Q(x_i)$ in $\mathbb{Q}[[x_1, \ldots, x_n]]$ and get a multiplicative sequence as YBL says in his answer, by taking $1 + \sum_j K_j = \prod_i Q(x_i)$, where the $K_j$ are taken as polynomials in the elementary symmetric polynomials (ie the Chern classes).

But I don't see how you can do this without appealing to this decomposition (or indeed in a more general setting without any idea of Chern classes).

For example, in Characteristic Classes by Milnor and Stasheff, they do the same explanation as I did above, but when they get to the formal power series part it seems that they just assume $B = \Lambda[t]$ for some commutative ring $\Lambda$ and from there I lose track of what is happening; this gets me really quite confused as to what's going on.

share|improve this question
    
Sorry, how is B_f a ring? I'd believe it if the sums started at b_0, but not if the sums start at 1: I think that sums starting at 1 are a multiplicative subgroup of the ring whose additive group structure is given by the infinite product of the B^i. –  Theo Johnson-Freyd Nov 23 '09 at 22:33
    
Oops, yes, sorry. It is exactly as you said: B_f is just a multiplicative subgroup of a ring; this ring is defined by considering all sums starting with b_0 in B^0, not only with 1 which is the case for B_f. –  Sam Derbyshire Nov 23 '09 at 22:53
add comment

3 Answers

up vote 6 down vote accepted

I think the idea is that using the splitting principle everything reduces to the first Chern class of line bundles: Chern classes of a general bundle $E$ are symetric functions of $c_1(L_i)$ where $\bigoplus L_i = E$.

If $Q(z)$ is a power series with constant term 1, you can define $K_n$ by the formula: $$ \sum K_n(x_1,\ldots,x_n) = \prod Q(z_j) $$ where $x_i$ is the $i$-th elementary symetric function of the variables $z_j$. Hogomeneity corresponds to the fact that the $z_j$ have degree 1.

I think the statement that every multiplicative sequence comes from such a power series is only true in caracteristic 0. A multiplicative sequence with coefficients in $A$ corresponds to a ring homomorphism from the Lazard ring $\mathbb{L} = \Omega^*(pt)$ to $A$ that is to a formal group law $F(t_1,t_2) \in A[[t_1,t_2]]$. There is a natural action of power series $f(z)$ satistfying $f(0) = 0$ and $f'(0) = 1$ on the Lazard ring; just change of coordinate of formal group laws $(F^f)(t_1,t_2) := f^{-1}(F(f(t_1),f(t_2))$. Now in caracteristic zero, this action is simply transitive: every law is equivalent to the additive one $(t_1,t_2) \mapsto t_1+t_2$ because we can define the logarithm of a law by formally integrating an invariant differential. This should correspond to the fact that every multiplicative sequence is defined by a power series $Q(z) = z/f(z)$.

share|improve this answer
    
Ok, thanks. Looking in the case that interests me the most for the moment (Chern classes), the formula you give does give a good way of computing the K_n. In this case Q is a power series with rational coefficients, and the K_n are just polynomials in the elementary symmetric polynomials, so polynomials in the Chern classes. I was just mulling over the definitions and never got too involved in actual calculations; doing an explicit calculation of the first few terms of the Todd genus helped a lot. –  Sam Derbyshire Nov 23 '09 at 22:45
    
Although I must say that still leaves me confused about how to do this in general without appealing to Chern roots. Ideally it should just be purely algebraic; but I can't even think of what power series ring Q should be in... –  Sam Derbyshire Nov 24 '09 at 2:45
add comment

Ok, I was getting confused about something rather silly.

As YBL says, getting a multiplicative sequence from a formal power series does just depend on the formula

$$1 + \sum_n K_n(\sigma_1, \ldots, \sigma_n) = \prod_j Q(z_j).$$

This is an algebraic identity in $\Lambda[[z_1, z_2, \ldots]] \cong \Lambda[[\sigma_1, \sigma_2, \ldots]]$ (although some care must probably be taken about the number of variables here).

Now, given a graded algebra over $\Lambda$ like the cohomology $H^{2 \bullet}(X,\mathbb{Q})$, we get the same identity (with higher degrees ending up vanishing), and this can be interpreted directly as giving something in $\mathbb{Q}[[z_1, \ldots, z_n]]$ in terms of something in $\mathbb{Q}[[\sigma_1, \ldots, \sigma_n]]$, and hence in this case establishing a connection between Chern roots and Chern classes, as they are just the $\sigma_1, \ldots, \sigma_n$.

It is interesting, as YBL notes, to ask if every multiplicative sequence comes from a formal power series in this way. I would enjoy it if someone could elaborate anymore more on what YBL has said so far.

share|improve this answer
add comment

The solutions to your questions are in Lemmas 1.2.1 and 1.2.2 (page 10) of "Topological Methods in Algebraic Geometry", F. Hirzebruch. As you say and those Lemmas confirm, «there is an essentially unique way to associate such a multiplicative sequence to a formal power series, and that all multiplicative sequences come from this».

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.