# When and why are Adams operations "non-negative"?

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 $$C$$?

• 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. Nov 3, 2019 at 1:19
• 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. Nov 3, 2019 at 1:24
• "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! Nov 3, 2019 at 1:29
• 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. Nov 3, 2019 at 1:37
• 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. Nov 3, 2019 at 9:05

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.