Posted by: Steven Sam | July 23, 2009

## Lambda-rings

In this post, I want to discuss Grothendieck’s $\lambda$-rings and how they provide an abstract setting for Riemann–Roch formalism. The references I’ll be using are

• Donald Knutson, Lambda-Rings and the Representation Theory of the Symmetric Group
• William Fulton and Serge Lang, Riemann–Roch Algebra

The definition of a $\lambda$-ring is a bit technical, but it starts with a commutative ring R together with operations $\lambda^i \colon R \to R$ for all nonnegative integers i such that $\lambda^0(r) = 1$ and $\lambda^1(r) = r$ for all r in R together with some axioms. In particular, we should say what these lambda operations do to sums and products, and we might also want to know what compositions of them look like. To motivate these axioms, we’ll look at K-theory (where it originates).

Let X be a topological space, and consider the set of all vector bundles over X (topological, smooth, holomorphic, whatever you want). We define lambda operations using exterior powers: $\lambda^i(E) = \bigwedge^i E$. Of course, the set of vector bundles on X isn’t a ring, but the free Abelian group of isomorphism classes of vector bundles is a ring if we use tensor product as the multiplication. But we have to define exterior products on “negatives” of isomorphism classes of vector bundles. For actual vector bundles, we have the identity $\bigwedge^n(E \oplus F) = \bigoplus_{i=0}^n \bigwedge^i E \otimes \bigwedge^{n-i} F$,

so we can use this to extend to “negatives” and we make this an axiom for a general $\lambda$-ring:

(L1) $\lambda^i(r+s) = \sum_{j=0}^i \lambda^j(r) \lambda^{i-j}(s)$ for all r and s in R.

In fact, the above identity holds if we pass to the Grothendieck group $K_0(X)$ of vector bundles over X (add the relations $E'+E'' = E$ whenever we have a short exact sequence of the form $0 \to E' \to E \to E'' \to 0$) because in general, if E’ is an extension of E and F, then $\bigwedge^n E'$ has a filtration whose associated graded is the direct sum on the RHS above.

What about products? i.e., what should $\lambda(rs)$ be? The exterior power of a tensor product of two vector bundles has a rather complicated expression. Nonetheless, there exist integer valued polynomials $P_n$ in 2n variables such that $\bigwedge^n(E \otimes F) = P_n(E, \bigwedge^2 E, \dots, \bigwedge^n E, F, \bigwedge^2 F, \dots, \bigwedge^n F)$.

How do we get these polynomials? Let $X_1, \dots, X_i, \dots, Y_1 \dots, Y_i, \dots$ be algebraically independent variables, and let $E_i, F_i$ denote the ith elementary symmetric function in the Xs and Ys, respectively. Then we define the polynomials $P_n$ via the identity $\displaystyle \sum_{n \ge 0} P_n(E_1, \dots, E_n, F_1, \dots, F_n) T^n = \prod_{i,j \ge 1} (1+X_iY_jT)$.

So we have some complicated family of polynomials, and the axiom

(L2) $\lambda^n(rs) = P_n(\lambda^1(r), \dots, \lambda^n(r), \lambda^1(s), \dots, \lambda^n(s))$ for all r and s in R.

There are also some integer-valued polynomials $P_{nm}$ of degree nm for expressing the compositions $\bigwedge^n \bigwedge^m E$. I won’t get into that, but this gives the third axiom

(L3) $\lambda^n(\lambda^m(r)) = P_{nm}(\lambda^1(r), \dots, \lambda^{nm}(r))$ for all r in R.

A morphism of $\lambda$-rings is a ring homomorphism which commutes with the lambda-operations.

For a simple combinatorial example, take X to be a point. In this case, vector bundles are just vector spaces, and $K_0(X) = \mathbf{Z}$. Identifying vector spaces with their dimension, the lambda operations become $\lambda^k(n) = \frac{n(n-1)\cdots (n-k+1)}{k!}$, which is just the binomial coefficient $\binom{n}{k}$ when n is nonnegative. A morphism $\varepsilon \colon R \to \mathbf{Z}$ is called an augmentation. For vector bundles, this map is given by sending a vector bundle to its rank and extending linearly to virtual vector bundles. We’ll assume our $\lambda$-rings are augmented.

We can place some further requirements and operations on our $\lambda$-rings. First, in $K_0(X)$, we naturally have a notion of what it means to be “positive”: any class which represents an actual vector bundle. The set of positive elements has the property that it’s closed under addition and multiplication, and every element of $K_0(X)$ can be expressed as a difference of two positive elements. Furthermore, whenever x is positive, $\lambda^i(x) = 0$ for sufficiently large i, and all positive elements of augmentation 1 (line bundles) have multiplicative inverses. We’ll take all of these features to be an axiom system for a “positive subset” of a $\lambda$-ring. Motivated by the K-theory, we’ll call positive elements of augmentation 1 line elements. We’ll assume that our $\lambda$-rings are equipped with a positive structure.

K-theory also has a nice involution: send a vector bundle to its dual bundle. In general, we’ll say that $x \mapsto x^\vee$ is an involution of our $\lambda$-ring if it satisfies

• $(x^\vee)^\vee = x$ for all x,
• $\varepsilon(x^\vee) = \varepsilon(x)$ for all x,
• $u^\vee = u^{-1}$ for all line elements u.

We’ll further assume that our $\lambda$-rings are equipped with an involution.

Another example of $\lambda$-rings comes from representation rings. Given a group G and a representation V, we define $\lambda^i(V)$ to be the representation $\bigwedge^i V$ with the diagonal action of G. The augmentation here sends a representation to its dimension (over the ground field), the positive elements are the representations, and the involution sends a representation to its dual. One particular example is when G is the general linear group $\mathbf{GL}_n$ and we consider only rational representations, so that the representation ring is the ring of symmetric functions in n variables (together with a multiplicative inverse for the product of the n variables). In this case, the lambda operations are plethysm: $\lambda^n(p) = e_n \circ p$ where $e_n$ is the nth elementary symmetric function, and positive means Schur positive. [I first saw $\lambda$-rings in context of symmetric functions, so the definitions seemed a bit mysterious to me.]

In the next post, I’ll discuss abstract Chern classes in the context of $\lambda$-rings and Riemann–Roch formalism, and say how this relates to Grothendieck–Riemann–Roch for proper maps between nonsingular varieties.

-Steven

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## Responses

1. If we’re in a symmetric tensor category, and take the Grothendieck group, do we get a $\lambda$-ring? Conversely, given a $\lambda$-ring, is it possible to derive a corresponding symmetric tensor category?

2. I think we at least want the category to be Abelian so that one can define exterior products (we need to know how to subtract maps and how to take images / cokernels). Do you know a way to define exterior products without this assumption? But, yes, given an Abelian symmetric tensor category, its Grothendieck group will be a $\lambda$-ring with a positive structure. I’m not sure how to get an augmentation or involution (maybe you need a forgetful functor to vector spaces).

As for your second question, probably: given a set S, we can construct a free $\lambda$-ring on S, which will be isomorphic to a polynomial ring over Z in variables s(i) for s in S and i a positive integer. In this case, $\lambda^i(s) = s(i)$. You can find the actual construction in Knutson’s book. This will also be the Grothendieck ring of the free Abelian symmetric tensor category C(S) generated by S. Every $\lambda$-ring is a quotient of such a free $\lambda$-ring, so we just add the appropriate exact sequences to C(S) to get the right Grothendieck group. This all sounds plausible to me, but I haven’t worked out any details, which is why I said “probably.”

There’s some more info here: http://ncatlab.org/nlab/show/Lambda-ring

3. Thanks for the response. I actually meant to implicitly include “abelian” in the definition of “symmetric tensor category,” which was probably a bad idea, since there seem to be different definitions depending on the context.

In any case, though, if the category is also rigid, then we should get an involution. This would include both the examples you mentioned.

Actually, I’m also curious- can we define exterior products in a symmetric tensor abelian $k$-linear category for some field $k$, when $k$ isn’t of characteristic zero? The only way I know to define the alternating square of $V$, say, is to take the cokernel of the map $V \otimes V \to V \otimes V$ which is the identity plus the braiding isomorphism. This should generalize to higher exterior powers, although one would have to quotient out by the sum of several different images. However, it isn’t right in charactersitic 2.

In the case of vector spaces, if I understand correctly, one wants to mod out $V \otimes V$ by the subspace generated $x \otimes x$. But the “diagonal map” is not linear and cannot be expected to be a morphism. Is there a way around this?

4. I think you can get around this issue in arbitrary characteristic by defining exterior powers as images of certain antisymmetrizing maps. Symmetric powers could still be cokernels, and we would also have divided powers as images.

5. One thing I do not really understand about the axioms for a positive structure (as given in Fulton/Lang): Can we actually prove (using the above axioms) that the multiplicative inverse of a line element is a line element as well? It is easy to see that this inverse exists, and if it is positive, it must be a line element, but I don’t see how to show that it is positive. Fulton/Lang call this obvious (or trivial). Of course, this *is* obvious in every application of lambda-rings I know, but out of curiosity I am still interested in the general case as well.

6. I think the idea is to use the polynomials $P_n$ above. Since $\lambda^i L = 0$ for i>1, and $\lambda^i 1 = 0$ for i>0, we can write $0 = \lambda^i L L^\vee = P_i(L, 0, \dots, 0, L^\vee, \lambda^2 L^\vee, \dots, \lambda^i L^\vee)$. I guess one can deduce that $\lambda^i L^\vee = 0$ for i>1 by induction on i and the relation above.

I haven’t checked this carefully, but I think that the polynomial $P_n(E_1, \dots, E_n, F_1, \dots, F_n)$ always has a term $E_1^n F_n$, and does not have a term of the form $E_1^n F_1^n$. Since $P_n$ is homogeneous of degree 2n (where $E_i, F_i$ have degree i), this is all that needs to be checked.

7. Ah, you are working over a special $\lambda$-ring. Then your proof actually simplifies: If $x$ is a line element and $y$ is an arbitrary element of a special $\lambda$-ring, then $\lambda_t\left(xy\right)=\lambda_{xt}\left(y\right)$, where $\lambda_t\left(u\right)$ denotes the formal power series $\sum_{i\geq 0}\lambda^i\left(u\right)t^u$ for every $u$ in the ring. This is even mentioned in Fulton/Lang. However, I am not sure whether Fulton/Lang assumed the $\lambda$-ring we are working in to be special; he even defined positive structure before defining special $\lambda$-rings. Then again, maybe positive structure implies speciality?

8. Anyway, both Fulton/Lang andCharles Weibel’s K-book state the above assertion (that the line elements form a subgroup of the $\lambda$-ring, particularly implying that the inverse of a line element is positive) without requiring it to be a special $\lambda$-ring (and before even defining what a special $\lambda$-ring is). Is this a mistake on their side or is there an argument why this always holds?

9. Should the wedge product $\wedge E’$ be written as $\wedge E$ in the sentence (below formula (L1) ): “… then $\wedge E’$ has a filtration whose associated graded is the direct sum on the RHS above.”?

10. Fei: No, E’ is an extension of E and F. It’s right as is, but it’s terrible notation probably.

11. Ah, I see. I was looking at the exact sequence $0\to E’\to E\to E”\to 0$. I though the $E’$ is the guy in this exact sequence.

12. After two years I just wanted to add that Fulton/Lang’s questionable claim (that the multiplicative inverse of a line element is a line element as well) is incorrect. See the footnote in the definition of a “positive structure” in my lambda-rings notes ( https://sites.google.com/site/darijgrinberg/lambda/lambda.pdf?attredirects=0&d=1 ); at the moment, this is footnote 29 on page 64 (continued onto page 65). The counterexample is actually even a special lambda-ring. The proof by Steven Sam above (for special lambda-rings) only shows that the inverse is 1-dimensional, but not that it is positive.