5
$\begingroup$

I wonder if there is a nice and short proof that the $K$-theory of a topological space is a special $\lambda$-ring without invoking the splitting principle and alike. Is it possible to show directly that $\lambda^k(V \otimes W)$ and $P_k(\lambda^1(V),...,\lambda^k(V),\lambda^1(W),...,\lambda^k(W))$ are stabily equivalent, without making unnatural choices? The same question for $\lambda^i(\lambda^k(V))$ and $P_{ik}(\lambda^1(V),...,\lambda^{ik}(V))$. It would be nice if there is some natural isomorphism on the level of vector spaces, which then may be glued to an isomorphism between vector bundles, perhaps with some extra summands on both sides which vanish in K-theory. Note that an affirmative answer would, in particular, answer this question by Darij Grinberg.

I hope that my question is clear enough although it is not precise. On the other hand, there is a precise generalization: Let $A$ be a topological ring, perhaps a Banach algebra. Is the Grothendieck ring of topological $A$-module bundles over a fixed $X$ a special $\lambda$-ring?

Improve this question
$\endgroup$
3
  • 2
    $\begingroup$ In your last paragraph, I don't think that the relevant Grothendieck group even has a natural ring structure without further assumptions. $\endgroup$
    – Neil Strickland
    Commented Sep 15, 2011 at 16:15
  • $\begingroup$ @Neil All I can think of is that tensor products might be 'wrong' without assumption (something like $A$-nuclearity, whatever that means). Or restrict to finitely generated topological $A$-module bundles etc. $\endgroup$
    – David Roberts ♦
    Commented Sep 15, 2011 at 23:40
  • $\begingroup$ Is a natural isomorphism on the level of vector space the same thing as an isomorphism in $Rep(GL_n)$? $\endgroup$
    – Will Sawin
    Commented Mar 9, 2013 at 23:48

1 Answer 1

Reset to default
1
$\begingroup$

All vector bundles are pull backs from the appropriate infinite Grassmanians/ classifying spaces. So it is sufficient to answer the question there.

But the category of vector bundles on such a space is just the category of representations of the appropriate group. So we just need to prove these for the special cases where the category is the category of representations of $GL_a \times GL_b$, and $V$ and $W$ are the standard representations of $GL_a$ and $GL_b$ respectively.

But the representation ring of those groups are well-known to be polynomial algebras generated by the $\lambda^i$s! So there is some polynomial formula. What polynomials must these be? By pulling back the representations to the representation ring of the maximal torus, we see that these polynomials must satisfy the identities that are usually taken to define them.

We can see immediately that this proof gives a natural isomorphism on the level of vector spaces, in that it gives a natural transformation between two endofunctors of the category of vector spaces where maps are isomorphisms, alias, an isomorphism between representations of $GL_n$.

Improve this answer
$\endgroup$
8
  • $\begingroup$ What characteristic are we in? I know this argument (or at least one that sounds like this -- I don't recall it ever saying anything geometric) being used in characteristic $0$, but I'm not sure how well it applies beyond that case. $\endgroup$
    – darij grinberg
    Commented Mar 10, 2013 at 22:37
  • $\begingroup$ I also can't really say I understand the last paragraph, about how you get a natural isomorphism. (Notice that the polynomials $P_k$ and $P_{ik}$ don't always have nonnegative coefficients, so one has to shovel some of the addends onto the other side, potentially obscuring the combinatorics.) $\endgroup$
    – darij grinberg
    Commented Mar 10, 2013 at 22:38
  • $\begingroup$ 1. It's characteristic $0$. $\lambda^k$ is not even defined in characteristic $p$, I don't think. 2. Yes, you have to shuffle addends to the other side. All I am doing is interpreting a natural isomorphism as a natural transformation, Clearly the $\lambda$-operations must be interpreted in this context as functors $Vect \to Vect$. To study vector bundles we only need to consider vector bundles up to isomorphism, so we only need the source category to consist of $n$-dimensional vector spaces, say. But the category of $n$-dimensional vector spaces is equivalent to the group $GL_n$. $\endgroup$
    – Will Sawin
    Commented Mar 10, 2013 at 22:57
  • $\begingroup$ I'm pretty sure $\lambda^k$ is defined in characteristic $p$ (as the exterior power). $\endgroup$
    – darij grinberg
    Commented Mar 11, 2013 at 6:18
  • $\begingroup$ "But the category of $n$-dimensional vector spaces is equivalent to the group $GL_n$." You mean the category of representations of the group $GL_n$? Sorry, I don't think so. And if I restrict myself to dimension $n$, my isomorphisms will only be canonical w.r.t. isomorphism of representations, not w.r.t. homomorphism of representations... $\endgroup$
    – darij grinberg
    Commented Mar 11, 2013 at 6:20

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .