6
$\begingroup$

We can think of the unary operations in a lambda-ring as integer linear combinations of Young diagrams; for example the operation $\lambda^n$ corresponds to the Young diagram with $n$ rows and one column.

Some of these operations are manifestly non-negative in the following sense: they're linear combinations of Young diagrams with natural number coefficients.

If we apply a manifestly non-negative operation to an element of the representation ring $R(G)$ coming from a representation of $G$, we get another element coming from a representation - not just a formal difference of such elements. Similarly, if we apply a manifestly non-negative operation to an element of the K-theory $K(X)$ coming from a vector bundle on $X$, we get another element coming from a vector bundle - not just a formal difference of such elements.

I'm confused about Adams operations. For $k > 1$, the Adams operation $\psi_k$ is apparently not manifestly non-negative, since it's given by an alternating sum of hook-shaped Young diagrams with $k$ boxes.

However, if we apply $\psi_k$ to an element of $K(X)$ coming from a vector bundle over $X$, I believe we get an element coming from a vector bundle. It's certainly true for line bundles: $\psi_k [L] = [L^{\otimes k}]$ when $[L]$ is the element of $K$-theory coming from a line bundle $L$. It's also true for bundles that split as a sum of line bundles, since $\psi_k : K(X) \to K(X)$ is a ring homomorphism. And I think it follows for all vector bundles using the splitting principle for K-theory (Corollary 4.3.4 here).

So, it seems that the Adams operations, while not manifestly non-negative, are still non-negative in the sense that they send elements of $K(X)$) coming vector bundles to elements coming from vector bundles - not merely formal differences of such.

My questions are:

1) Is this true?

2) If so, which integer linear combinations of Young diagrams give operations that are non-negative in this sense?

3) What's really going on here? In particular, I've defined "non-negative" using $K(X)$, but these should be examples of a more general phenomenon. The Grothendieck ring $K(C)$ of any symmetric monoidal Cauchy-complete $\mathbb{Q}$-linear category $C$ is a lambda-ring, in a way that generalizes this. We can define "non-negative" operations on $K(C)$ to be those sending elements coming from objects of $C$ to elements coming from objects of $C$. Are Adams operations always non-negative on $K(C)$, or this just true for certain $C$? Which $C$ are these? And which linear combinations of Young diagrams give operations that are non-negative on $K(C)$ for all symmetric monoidal Cauchy-complete $C$?

$\endgroup$
8
  • 4
    $\begingroup$ For $\rho$ a representation, $\rho(g^k)$ is not a representation unless $G$ is a abelian. No operation preserves non negativity for all groups unless it is a nonnegative combination of Schur functors, as you can see by plugging in the standard representation of $GL_n$. An operation preserves non negativity in the representation ring of all abelian groups if and only if the associated symmetric polynomial has nonnegative coefficients. $\endgroup$
    – Will Sawin
    Nov 3, 2019 at 1:19
  • 1
    $\begingroup$ There might be many examples of operations that are only nonnegative for symmetric monoidal abelian categories of a certain type. Simple Lie groups with no automorphisms of their Dynkin diagram have only self-dual representations, making $\operatorname{Sym}^2 V + \wedge^2 V -1$ nonnegative, at least when $V$ is nonzero. $\endgroup$
    – Will Sawin
    Nov 3, 2019 at 1:24
  • 1
    $\begingroup$ "For $\rho$ a representation, $\rho(g^k)$ is not a representation unless $G$ is abelian." Oh, duh. I guess this somehow analogous to the splitting principle then: people study Adams operations on $\mathrm{Rep}(G)$ with $G$ a compact Lie group by looking at what they do to $\mathrm{Rep}(T)$ when $T$ is a maximal torus, where the formula I gave actually makes sense! $\endgroup$
    – John Baez
    Nov 3, 2019 at 1:29
  • $\begingroup$ Given the mistake that Will Sawin caught, I'm gonna delete the stuff about representation rings in my original post. Nonetheless I'm happy to hear about which lambda-operations are nonnegative in which representation rings. $\endgroup$
    – John Baez
    Nov 3, 2019 at 1:37
  • 2
    $\begingroup$ Just for clarity I think the splitting principle argument is also wrong. The map being objective on $K$-theory implies almost nothing about what it does on vector bundle classes. Probably the tangent bundle to $\mathbb P^2$ provides a counterexample. $\endgroup$
    – Will Sawin
    Nov 3, 2019 at 9:05

1 Answer 1

4
$\begingroup$

The Adams operations aren't always non-negative. (They are if you restrict to the subring generated by line bundles, though.)

Here's a counterexample in the world of finite CW-complexes. Let $X$ be the truncated projective space $\mathbb{RP}^{2n-1}/\mathbb{RP}^{2t-1}$. It's reduced $K$-theory was calculated by Adams here to be $\mathbb{Z}\oplus\mathbb{Z}/2^{n-t}\mathbb{Z}$. Let $\mu$ and $\nu$ be generators of each factor. They are best understood as follows. The inclusion of the $2t$-skeleton gives a map $S^{2t}\hookrightarrow X$. Pulling back along this map sends $\mu$ to a generator of $\tilde{K}(S^{2t})$. Let $\eta$ be the complexified tautological line bundle over $\mathbb{RP}^{2n-1}$. Pulling back along the quotient map $\mathbb{RP}^{2n-1}\rightarrow X$ sends $\nu$ to the class $([\eta]-1)^{t+1}=(-2)^{t}([\eta]-1)$. Note that the last equality is a simple consequence of the fact that real line bundles are self dual. Adams also calculated the effect of the Adams operations on these generators. For example, if $k$ is odd then $$\psi^{k}\mu=k^{t}\mu+\frac{k^{t}-1}{2}\nu$$ (Sidenote: that 2 in the denominator is the poison dart in Adams' vector-fields-on-spheres proof. For our purposes it is only important that $\psi^{k}\mu$ has a component in both the $\mu$ and $\nu$ factor. If we add a $t$ dimensional trivial bundle to the class $\mu$, then it is represented by an honest bundle $V$, which can be constructed via clutching, by cutting apart the $S^{2t}$ that comprises the $2t$-skeleton of $X$. Therefore, $$\psi^{k}[V]=k^{t}[V]+ t -k^{t}t +\frac{k^{t}-1}{2}[\eta] -\frac{k^{t}-1}{2}$$ Multiplying the clutching function of $V$ by $k^{t}$ (as an element in $\pi_{2t-1}U(t)=\mathbb{Z}$ gives a vector bundle $V_{k^{t}}$ representing $k^{t}\mu+t$, so $$\psi^{k}[V]=[V_{k^{t}}]+\frac{k^{t}-1}{2}[\eta] -\frac{k^{t}-1}{2}$$ Now, it's a general fact that if $E$ is a vector bundle such that the class $[E]-n$ is representable by a vector bundle, then the last $n$ Chern classes of $E$ must be zero. We can achieve this obstruction in our example by choosing the right $k$, $n$, and $t$. The first term has a single non-vanishing (top) chern class, which is $k^{t}$ times a generator of $H^{2t}(X)=\mathbb{Z}$, and the chern class of the second term is easily understood in terms of the chern class of $\eta$. In particular its top chern class will be nonzero as long as $n$ is very large.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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