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 $C$?

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    $\begingroup$ 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 '19 at 1:19
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    $\begingroup$ 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 '19 at 1:24
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    $\begingroup$ "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 '19 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 '19 at 1:37
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    $\begingroup$ 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 '19 at 9:05

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.


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