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Given a commutative ring $A$ with unity, Grothendieck used universal polynomials to define a special $\lambda$-ring structure on $\Lambda(A):=1+t\:A[[t]]$. Suppose $A$ is graded, say $A=\bigoplus_{i=o}^\infty A_i$. In Riemann-Roch Algebra, p. 11, Fulton and Lang define $\Lambda^{\circ}(A):=\{1+a_1t+a_2t^2\ldots\mid a_i\in A_i\}$. Then on page 15 they state that since the product and $\lambda$ operations of $\Lambda(A)$ take $\Lambda^\circ(A)$ to itself, $\Lambda^\circ(A)$ becomes a $\lambda$-ring (without unit). They use this $\lambda$-ring structure of $\Lambda^\circ(A)$ in proof of Theorem 3.1 on p. 16.

However, a straightforward computation shows that the product in $\Lambda(A)$ does not take $\Lambda^\circ(A)$ to itself. For example, if $1+a_1t+a_2t^2\ldots$ and $1+b_1t+b_2t^2\ldots$ are elements in $\Lambda^\circ(A)$, then their product using the product of $\Lambda(A)$ is given by $1+P_1(a_1;b_1)t+P_2(a_1,a_2;b_1,b_2)t^2+\ldots$, where $P_1,P_2,\ldots$ are certain universal polynomials. But $P_1(a_1;b_1)$ turns out to be $a_1b_1$ (see here, p. 22) and $a_1b_1$ is not in $A_1$, which shows the product is not in $\Lambda^\circ(A)$.

Question. Is there an error in the book? If yes, can it be fixed?

Edit. If you know other errors in this book that one should be aware of, please share it here.

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Another mistake in Fulton-Lang's book is the assertion that the $j$-th Bott element of a positive element in a $\lambda$-ring $R$ is invertible in $R[1/j]$; see p. 24 (this is only true if $R$ is augmented with a locally nilpotent $\gamma$-filtration). –  Damian Rössler Sep 30 '12 at 7:59
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@Damian: Thank you! Your comment gave me the idea to ask about other known errors in the book. –  Mahdi Majidi-Zolbanin Sep 30 '12 at 14:38
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4 Answers 4

up vote 5 down vote accepted

As others have said, the definition of the Chern ring there is wrong. But if memory serves, the only mistake is that they forgot to introduce the right multiplication law on the sets of power series they consider. The usual one in the theory is given by the universal formulas for exterior powers of tensor products $\Lambda^n(E\otimes F)$, but the one they want is for Chern classes $c_n(E\otimes F)$. When $n=1$ and $E$ and $F$ are line bundles, the first is multiplication and the second is addition. So it's obviously just an oversight, but one that can be confusing if you're seeing these things for the first time. (In the copy at U Chicago, someone mercifully added a warning note in the margin. There are a few obvious suspects.)

If you want a reference where the details are correct, I'd recommend SGA6. Grothendieck's introduction in expose 0 is very clear. Page 28 is where the discussion of the Chern ring starts. If I remember, Berthelot's expose goes into more depth, but I found Grothendieck's easier to read. Berthelot gets to the Chern ring on page 344. Atiyah-Tall is also generally a good reference, but I think they don't cover the Chern ring (although they do introduce the gamma-filtration).

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Dear James: Are you saying that one can correct the issue by using a different set of universal polynomials to define multiplication in the graded ring that they consider? –  Mahdi Majidi-Zolbanin Sep 30 '12 at 5:26
    
Yes, I was saying that and also that the universal polynomials are the ones that describe the Chern classes of tensor products. Note that the polynomials are not completely universal -- they do depend on the ranks of the factors. The usual way around this is to assume both factors have rank zero (which is why you get non-unital rings). –  JBorger Sep 30 '12 at 6:09
    
What is the definition of universal as in universal polynomials? I mean I know how they are defined, but why are they called universal? Does it simply mean that the polynomials have integral coefficients or is it more to it? –  Mahdi Majidi-Zolbanin Oct 7 '12 at 23:14
    
'Universal' just emphasizes that the polynomial is independent of the ring. For instance, the Leibniz rule $d(xy)=xdy+ydx$ for derivations can be expressed as $d(xy)=P(x,y,dx,dy)$, where $P(a,b,c,d)=ad+bc$, and the polynomial $P$ is independent of the ring on which you have a derivation. You can imagine other kind of operators where $d(xy)$ is a polynomial in $x,y,dx,dy$ but that polynomial can depend on the ring. –  JBorger Oct 8 '12 at 1:21
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Hazewinkel in http://arxiv.org/pdf/0804.3888.pdf warns about an error on page 15, second paragraph of thie book. In fact he advises to "steer clear" of the book!

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Predictably, when I try to consult Google Books to find the gaffe, "pages 15 and 16 are not shown". –  Todd Trimble Sep 30 '12 at 1:39
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I see! He also mentions the book review by K. R. Coombes in Math. Rev. (88h:14011). I should read the review to see what Coombes wrote about the book. Thanks! –  Mahdi Majidi-Zolbanin Sep 30 '12 at 2:07
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This is not an answer, as I don't exactly know what Fulton and Lang are trying to achieve with the $\lambda$-ring structure on $\Lambda^{\circ}\left(A\right)$ (I must admit that, while I had the quixotic intent to read and rewrite Fulton-Lang's Chapter I in the notes that you cited, I never found the resolve to walk that talk). I can confirm your counterexample.

What I think can be done (don't know if it is of any help) is the following: For every $i\in\mathbb N$, let $\Lambda^{i}_{\circ}\left(A\right)$ be the subset of $\Lambda\left(A\right)$ consisting of all formal power series of the form $1+a_1t+a_2t^2+a_3t^3+...$ with every $k$ satisfying $a_k\in A^{ik}$. Then, each such $\Lambda^{i} _ {\circ}\left(A\right)$ is an additive subgroup of $\Lambda\left(A\right)$, and the direct sum $\bigoplus\limits_{i\in\mathbb N}\Lambda^{i}_{\circ}\left(A\right)$ is well-defined and a sub-$\lambda$-ring of $\Lambda\left(A\right)$. (This is easy to prove by means of the usual grading on the ring of symmetric functions.) This sub-$\lambda$-ring, of course, is graded (and does have a $1$). I have no idea in how far it is what Fulton and Lang wanted.

We could also construct a greater graded sub-$\lambda$-ring of $\Lambda\left(A\right)$ by allowing $i$ rational (with $A^x$ defined as $0$ when $x\not\in\mathbb Z$), but then it will be graded by rationals. This greater graded sub-$\lambda$-ring is actually dense in $\Lambda\left(A\right)$ (in the usual topology on formal power series).

Does it make sense to replace $\Lambda^{\circ}\left(A\right)$ by $\Lambda^{\geq 1}_{\circ}\left(A\right)$ in the definition of a Chern class homomorphism? I don't know. It seems that most notions in Fulton-Lang are motivated by geometry, and without understanding it I am not the one to judge.

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Dear Darij: Thank you for your suggestion. Are you sure $1+t$ (the $1$ of $\Lambda(A)$) is in $\bigoplus_{i\in\mathbb{N}}\Lambda^i_\circ(A)$? In any case, I think Fulton & Lang wanted a $\lambda$-ring structure on $\Lambda^\circ(A)$ to prove a Splitting Principle for abstract Chern classes (Theorem 3.1), which in the geometric case is evident. But even with your suggestion, I am not sure if it is possible to fix the proof of Theorem 3.1, because I believe the proof has other issues, e.g., I think they mix up the notion of $\lambda$-homomorphism introduced on page 15 with $\lambda$-ring homomo. –  Mahdi Majidi-Zolbanin Sep 30 '12 at 5:22
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$1+t$ lies in $\Lambda^0_{\circ}\left(A\right)$, the $0$-th graded component of $\Lambda_{\circ}\left(A\right)$. Anyway, I guess James Borger understands better what Fulton and Lang were trying to say. –  darij grinberg Sep 30 '12 at 14:53
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I'm not sure what product you are thinking of on $\Lambda^0(A)$, but the one I'm thinking of, and the one that I believe is implicitly used in Fulton-Lang is the usual product on power series. So in particular, $(1+a_1 t+\cdots)\cdot (1+b_1 t+\cdots) = 1 + (a_1 + b_1)t + \cdots$.

There is no problem of grading.

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This product is not the product, but the sum of $\Lambda\left(A\right)$. –  darij grinberg Sep 29 '12 at 17:04
    
As Darij wrote, in the λ-ring structure the ordinary product that you wrote above is considered the addition operation. So what you did here is you added the elements. Multiplication is given by universal polynomials –  Mahdi Majidi-Zolbanin Sep 29 '12 at 17:07
    
Yes, you are right, sorry for this too prompt answer. –  Baptiste Calmès Oct 4 '12 at 4:51
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