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Let $E$ and $F$ be vector bundles on a smooth projective variety, say.

A. Lascoux ("Classes de Chern d'un produit tensoriel", C. R. Acad. Sci. Paris Sér. A-B 286 (1978), no. 8, A385–A387) gave formulas for the Chern classes of $E \otimes F$, $Sym^2 E$ and $\bigwedge^2 E$ in terms of the Chern classes of $E$ and $F$.

Unfortunately, I don't have access to Lascoux's article, and a bit of Googling didn't find them reproduced elsewhere.

Does anyone know another reference (preferably freely available online) where these formulas are written down?

Edit: In the comments, Robert Bryant suggests Hirzebruch's book "Topological methods in algebraic geometry" as a reference. Indeed, there is a formula there for the generating function for the Chern classes of $E \otimes F$and $\bigwedge^p E$, namely the obvious thing you get from the splitting principle and Whitney sum formula. So this gives you some answer in terms of Chern roots of $E$ and $F$.

But he content of Lascoux's formulas (I guess; I mean, I haven't actually seen them) is to rearrange this into an expression just in terms of the Chern classes of $E$ and $F$. I can probably (he claimed) do this in whatever case I care about, but the real intent of my question is actually to get a reference.

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    $\begingroup$ They are certainly in Hirzebruch's book, "Topological methods in algebraic geometry", but I don't imagine that this is online. However, these formulae are very easy to derive using the splitting principle and generating functions, so it shouldn't be any trouble to derive them if you need them: Just assume, without loss of generality, that $E$ and $F$ are sums of line bundles, use the fact that $c_1(E\otimes F) = c_1(E)+c_1(F)$ when they are line bundles, and then use the Whitney sum formula that $c(E\oplus F) = c(E)c(F)$ when $c(E)$ is the total Chern class of $E$. Combinatorics does the rest. $\endgroup$
    – Robert Bryant
    Commented Mar 8, 2013 at 0:30
  • $\begingroup$ Moreover, if you're willing to work over $\mathbb C$ (or do some tricks), you can prove such formulas once and for all on the classifying space $BGL(\dim E) \times BGL(\dim F)$ of a pair of bundles. This gives a sense of why there should be a universal formula. $\endgroup$
    – Allen Knutson
    Commented Mar 8, 2013 at 3:03
  • $\begingroup$ Thanks for the comments. Just to clarify, I'm familiar with the splitting principle and so on; part of my question was asking about the content of "Combinatorics does the rest." Moreover, I'm really looking for an actual reference, rather than an explanation. @Robert Bryant: I'll look in Hirzebruch's book, if my library has it. I'll still be interested to hear of other references! $\endgroup$
    – user5117
    Commented Mar 8, 2013 at 16:19
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    $\begingroup$ @Artie: The formulae you want are in Hirzebruch's book. They are given in Theorem 4.4.3, on page 64 of the English translation. My 'combinatorics does the rest' statement is just the observation that the formulae are just the result expressing everything in terms of elementary symmetric functions using the splitting principle and the converting the resulting polynomials in to polynomials in the two groups of elementary symmetric functions. It really is elementary, as you'll see when you look in Hirzebruch. (He doesn't actually do the combinatorics, but it's not hard.) $\endgroup$
    – Robert Bryant
    Commented Mar 8, 2013 at 20:46
  • $\begingroup$ @Robert Bryant: Thanks for the more detailed reference; Theorem 4.4.3 is indeed what I was referring to in my edit above. Anyway, maybe I was being overly fussy in wanting an actual reproduction of the formulas. $\endgroup$
    – user5117
    Commented Mar 9, 2013 at 15:49

2 Answers 2

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You say you don't have access to Lascoux's paper, but it is actually available online at the BNF:

http://gallica.bnf.fr/ark:/12148/bpt6k62341359/f397.image

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  • $\begingroup$ Thanks! This seems to be one of those things that Google is not good at finding. $\endgroup$
    – user5117
    Commented Mar 9, 2013 at 16:00
  • $\begingroup$ The formula provided in this work as well as more recent one, arxiv.org/pdf/1012.0014‎, are rather difficult. How ever I am interested in the top Chern class and I was wondering if you guys can write a somehow explicit formula for top Chern class !! Thanks in advance. $\endgroup$
    – Mohammad Farajzadeh-Tehrani
    Commented Dec 23, 2013 at 1:34
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The splitting principle gives the expression $c(E\otimes F)=\prod_x\prod_y (1+x+y)$ that combinatorics has to write in terms of symmetric functions. To argue that this is a necessary step, one could say that Macdonald's book ``Symmetric functions and Hall polynomials'' is mostly devoted to computing symmetric functions without using any variable. As a matter of fact, the Chern classes of a tensor product are given p.67 in Macdonald's book. To illustrate that $\prod_x\prod_y (1+x+y)$ is not an appropriate expression for a geometer, one can take for example the problem of finding a criterium for numerical positivity of ample vector bundles. Positivity of Chern classes is not enough. But the criterium must be compatible with tensor products. The fact that $\prod_x\prod_y (1+x+y)$ expands positively in terms of Schur functions forces to add positivity of Schur functions to the criterium. In fact, this criterium is now sufficient, as proved by Lazarfeld and Fulton (Annals of Math 118(1983) 35-60). The coefficients of the expansion in the Schur basis of $c(E\otimes F)$ are determinants of binomial coefficients which can be studied by using non-intersecting paths, after Gessel and Viennot. Moreover they are equal to specializations of Schubert polynomials in $x_1=1=x_2\ldots$, $y_1=0=y_2\ldots$ because Schubert polynomials satisfy a Cauchy formula (cf. A.Lascoux, Symmetric functions & Combinatorial operators on polynomials, CBMS/AMS Lectures Notes ${\bf 99}$, (2003), p. 161.

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