## May 26, 2009

The Adams operations in K-theory are supposed to be defined by three properties

1. $\psi_k: K(X)\to K(X)$ is a ring-homomorphism.
2. $\psi_k\circ \psi_l = \psi_{l k}$.
3. Acting on a line bundle, $L$, $\psi_k(L) = L^{k}$.

Between that, and the splitting principle, one is supposed to be able to crank them out, explicitly, whenever one needs them.

That’s a pain.

So, in the interest of providing myself (and anyone else who might find it useful) with a handy online reference, here are the first few Adams operations, acting on a vector bundle, $E$.

\begin{aligned} \psi_1(E) &= E\\ \psi_2(E) &= \mathop{Sym}^2(E) - \wedge^2(E)\\ &=\underset{\color{brown}\frac{n(n+1)}{2}}{\array{\begin{svg}\end{svg}}} - \underset{\color{brown}\frac{n(n-1)}{2}}{\array{\begin{svg}\end{svg}}}\\ \psi_3(E) &= \underset{\color{brown}\frac{n(n+1)(n+2)}{3!}}{\array{\begin{svg}\end{svg}}} - \underset{\color{brown}\frac{(n-1)n(n+1)}{3}}{\array{\begin{svg}\end{svg}}}+ \underset{\color{brown}\frac{n(n-1)(n-2)}{3!}}{\array{\begin{svg}\end{svg}}}\\ \psi_4(E) &= \underset{\color{brown}\frac{n(n+1)(n+2)(n+3)}{4!}}{\array{\begin{svg}\end{svg}}} - \underset{\color{brown}\frac{(n-1)n(n+1)(n+2)}{8}}{\array{\begin{svg}\end{svg}}}+ \underset{\color{brown}\frac{(n+1)n(n-1)(n-2)}{8}}{\array{\begin{svg}\end{svg}}} - \underset{\color{brown}\frac{n(n-1)(n-2)(n-3)}{4!}}{\array{\begin{svg}\end{svg}}} \end{aligned}

where I’ve indicated the (anti)symmetrization of tensor powers of $E$ by the associated Young diagram. Below each Young diagram, I’ve indicated the rank of the resulting vector bundle, where $n=\mathop{rank}(E)$.

The relation $\psi_4 = \psi_2\circ\psi_2$ follows from

\begin{aligned} \mathop{Sym}^2\left(\array{\begin{svg}\end{svg}}\right) &= \array{\begin{svg}\end{svg}} + \underset{\color{brown}\frac{n^2(n^2-1)}{12}}{\array{\begin{svg}\end{svg}}} \\ \wedge^2\left(\array{\begin{svg}\end{svg}}\right) &= \array{\begin{svg}\end{svg}}\\ \mathop{Sym}^2\left(\array{\begin{svg}\end{svg}}\right) &= \array{\begin{svg}\end{svg}} + \array{\begin{svg}\end{svg}}\\ \wedge^2\left(\array{\begin{svg}\end{svg}}\right) &= \array{\begin{svg}\end{svg}} \end{aligned} and linearity under direct sums (and differences).

The proof, of all of the above, follows from a standard, boring, application of the splitting principle. But I rather wish the explicit formulæ were recorded somewhere readily accessible. Now they are …

#### Update (5/27/2009)

Hmmm…

Having written this much (and checking the $k=5$ case), it seems natural to conjecture that the kth Adams operation is the alternating sum of “hook representations” H^k_s(E) =\array{ \arrayopts{\align{baseline}\rowalign{bottom}} s\begin{cases}\space{30}{0}{0}\end{cases}\overset{k-s}{\overbrace{\array{ \begin{svg}\end{svg}}}}},\qquad \mathop{rank}(H^k_s(E)) = \frac{(n+k-s-1)!}{k(n-s-1)!(k-s-1)!s!} with $k$ boxes:

(1)$\psi_k(E) = \sum_{s=0}^{k-1} (-1)^s H^k_s(E)$

Anyone know if that’s true and, if so, where it’s proven?

#### Update (5/28/2009)

Actually, (1) is not so hard to prove. Volker has a sketch of a proof, below. Here’s another.

First let’s see that it behaves correctly under direct sums. By the splitting principle, we can work inductively on the rank of $E$, and check that it behaves correctly for the direct sum with a line bundle. If we denote $H^j_{-1}(E)\equiv \delta_{j,0}$, we can compactly write $H^k_s(E+L)= \sum_{j=1}^{k-s} H^{k-j}_{s-1}(E)\otimes L^j + \sum_{j=0}^{k-s-1} H_s^{k-j}(E)\otimes L^j$ Plugging this into the RHS of (1), we obtain $\sum_{s=0}^{k-1} (-1)^s H^k_s(E+L) = L^k + \sum_{s=0}^{k-1}(-1)^s H^k_s(E)$ as expected. Also, since $H^k_s(L \otimes E) = L^k\otimes H^k_s(E)$ (1) behaves correctly under tensor products, and thus is a ring homomorphism of $K(X)$.

I guess there was no point to this post, after all. Even I can remember (1): the kth Adams operation is the alternating sum of $k$-hooks!

#### Update (5/29/2009): Atiyah

Surely, this was known to the ancients. And, indeed, Dan Freed points me to an old paper of Atiyah (“Power Operations in K-Theory”, Quart. Jour. Math., 17 (1966) 165-193), where the connection with representations of the symmetric group are worked out.

In closing, I should point out a particularly nice feature of (1). Let $M$ be the $(k-1)$-dimensional “Standard” representation of $S_k$ (the quotient of the $k$-dimensional “permutation” representation by the 1-dimensional trivial representation). The $k$-hooks are the exterior powers of $M$. In other words, the RHS of (1) corresponds to the element $\wedge_{-1}(M) = \sum_s (-1)^s \wedge^s(M)$ of the representation ring of $S_k$.

Posted by distler at May 26, 2009 12:00 AM

TrackBack URL for this Entry:   http://golem.ph.utexas.edu/cgi-bin/MT-3.0/dxy-tb.fcgi/1976

If you don’t want to get your hands dirty with the splitting principle you can of course use

(1)$\sum_{i=1}^k \psi_i(E) \; \mathop{Sym}^{k-i}(E) = k\; \mathop{Sym}^k(E)$

for $k=1,\dots,n$ and solve for $\psi_n$. Though there remains an annoying amount of Young diagramatics to boil it down to your beautiful formulae…

Posted by: Volker Braun on May 26, 2009 6:03 AM | Permalink | Reply to this

There’s an explicit formula in Hatcher’s book Vector Bundles and K-theory in terms of Newton polynomials and exterior powers which is probably the same thing.

Posted by: Aaron Bergman on May 27, 2009 2:05 PM | Permalink | Reply to this

### Sums of hooks

Oooh nice equation! I’ve convinced myself that one can prove it with the recursion formula and Young gymnastics, but its painful…

I think the right point of view is this: We do have a ring homomorphism

(1)$\psi_k(E+1) = \psi_k(E)+1,$

where $1$ is the trivial line bundle. But naive dimensional reduction yields three terms:

1. all indices in $E$
2. one index in $1$, $k-1$ indices in $E$.
3. all indices in the trivial bundle $1$.

The first term is $\psi_k(E)$ and the third term is $\psi_k(1)=\mathop{Sym}^k(1)=1$. The second term must vanish.

In other words: $\psi_k$ is some expression in $k$-block young diagrams, corresponding to representations of the symmetric group $S_k$. If we restrict to $S_{k-1}$, this must vanish.

In Young diagram language, restricting to $S_{k-1}$ is the sum of all Young diagrams with one box removed. In particular, the $k$-hook decomposes into two $(k-1)$-hooks, the $k$-row into a $(k-1)$-row, and the $k$-column into a $(k-1)$-column. (How did you draw those diagrams in itex?)

Now its easy to see that

(2)$\psi_k(E) = \mathop{Sym}^k(E) + (\text{stuff}) = (k\text{-row}) + (\text{stuff}).$

The remaining stuff is determined by demanding that it vanishes as a $S_{k-1}$-rep, and you get the alternating sum of hooks.

Posted by: Volker Braun on May 28, 2009 6:06 AM | Permalink | Reply to this

### Re: Sums of hooks

We do have a ring homomorphism $\psi_k(E+1) = \psi_k(E) + 1$ where $1$ is the trivial line bundle.

I think you need to check it for an arbitrary line bundle, not just the trivial one. But the rest of your argument seems good.

How did you draw those diagrams in itex?

$$\wedge^2\left(\array{\begin{svg} <svg xmlns="http://www.w3.org/2000/svg" xmlns:xlink='http://www.w3.org/1999/xlink' width="1.5em" height=".75em" viewBox="0 0 24 12"> <desc>Young Diagram with Dynkin indices (2,0)</desc> <g transform="translate(1,1)"> <use xlink:href="#rect123"/> <use transform="translate(10,0)" xlink:href="#rect123"/> </g> </svg> \end{svg}}\right) = \array{\begin{svg} <svg xmlns="http://www.w3.org/2000/svg" xmlns:xlink='http://www.w3.org/1999/xlink' width="2em" height="1.5em" viewBox="0 0 32 24"> <desc>Young Diagram with Dynkin indices (2,1,0,0)</desc> <g transform="translate(1,3)"> <use xlink:href="#rect123"/> <use transform="translate(10,0)" xlink:href="#rect123"/> <use transform="translate(20,0)" xlink:href="#rect123"/> <use transform="translate(0,10)" xlink:href="#rect123"/> </g> </svg> \end{svg}}$$

(where the square was defined in the first such equation) produces

$\wedge^2\left(\array{\begin{svg}\end{svg}}\right) = \array{\begin{svg}\end{svg}}$

Posted by: Jacques Distler on May 28, 2009 2:00 PM | Permalink | PGP Sig | Reply to this

Now I’m curious too — since I’ve been pondering how Adams operations link what James Borger calls the ‘orthodox’ and ‘heterodox’ reasons for getting interested in $\lambda$-rings. You’re probably doing ‘orthodox’ stuff, since you’re talking about vector bundles. Borger is doing ‘unorthodox’ stuff: Frobenius lifts.