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May 26, 2009

Adams Operations

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

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

ψ 1(E) =E ψ 2(E) =Sym 2(E) 2(E) = Young Diagram with Dynkin indices (2,0) n(n+1)2 Young Diagram with Dynkin indices (0,1) n(n1)2 ψ 3(E) = Young Diagram with Dynkin indices (3,0,0) n(n+1)(n+2)3! Young Diagram with Dynkin indices (1,1,0) (n1)n(n+1)3+ Young Diagram with Dynkin indices (0,0,1) n(n1)(n2)3! ψ 4(E) = Young Diagram with Dynkin indices (4,0,0,0) n(n+1)(n+2)(n+3)4! Young Diagram with Dynkin indices (2,1,0,0) (n1)n(n+1)(n+2)8+ Young Diagram with Dynkin indices (1,0,1,0) (n+1)n(n1)(n2)8 Young Diagram with Dynkin indices (0,0,0,1) n(n1)(n2)(n3)4! \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}<svg xmlns="http://www.w3.org/2000/svg" xmlns:xlink='http://www.w3.org/1999/xlink' width="3.75em" height=".75em" viewBox="0 0 60 12"> <desc>Young Diagram with Dynkin indices (2,0)</desc> <g transform="translate(20,1)"> <rect id="rect123" fill="#FCC" stroke="#000" stroke-width="2" width="10" height="10"/> <use transform="translate(10,0)" xlink:href="#rect123"/> </g> </svg>\end{svg}}} - \underset{\color{brown}\frac{n(n-1)}{2}}{\array{\begin{svg}<svg xmlns="http://www.w3.org/2000/svg" xmlns:xlink='http://www.w3.org/1999/xlink' width=".75em" height="1.5em" viewBox="0 0 12 24"> <desc>Young Diagram with Dynkin indices (0,1)</desc> <g transform="translate(1,3)"> <use xlink:href="#rect123"/> <use transform="translate(0,10)" xlink:href="#rect123"/> </g> </svg>\end{svg}}}\\ \psi_3(E) &= \underset{\color{brown}\frac{n(n+1)(n+2)}{3!}}{\array{\begin{svg}<svg xmlns="http://www.w3.org/2000/svg" xmlns:xlink='http://www.w3.org/1999/xlink' width="2em" height=".75em" viewBox="0 0 32 12"> <desc>Young Diagram with Dynkin indices (3,0,0)</desc> <g transform="translate(1,1)"> <use xlink:href="#rect123"/> <use transform="translate(10,0)" xlink:href="#rect123"/> <use transform="translate(20,0)" xlink:href="#rect123"/> </g> </svg>\end{svg}}} - \underset{\color{brown}\frac{(n-1)n(n+1)}{3}}{\array{\begin{svg}<svg xmlns="http://www.w3.org/2000/svg" xmlns:xlink='http://www.w3.org/1999/xlink' width="1.5em" height="1.5em" viewBox="0 0 24 24"> <desc>Young Diagram with Dynkin indices (1,1,0)</desc> <g transform="translate(1,3)"> <use xlink:href="#rect123"/> <use transform="translate(10,0)" xlink:href="#rect123"/> <use transform="translate(0,10)" xlink:href="#rect123"/> </g> </svg>\end{svg}}}+ \underset{\color{brown}\frac{n(n-1)(n-2)}{3!}}{\array{\begin{svg}<svg xmlns="http://www.w3.org/2000/svg" xmlns:xlink='http://www.w3.org/1999/xlink' width=".75em" height="2em" viewBox="0 0 12 32"> <desc>Young Diagram with Dynkin indices (0,0,1)</desc> <g transform="translate(1,1)"> <use xlink:href="#rect123"/> <use transform="translate(0,10)" xlink:href="#rect123"/> <use transform="translate(0,20)" xlink:href="#rect123"/> </g> </svg>\end{svg}}}\\ \psi_4(E) &= \underset{\color{brown}\frac{n(n+1)(n+2)(n+3)}{4!}}{\array{\begin{svg}<svg xmlns="http://www.w3.org/2000/svg" xmlns:xlink='http://www.w3.org/1999/xlink' width="2.75em" height=".75em" viewBox="0 0 44 12"> <desc>Young Diagram with Dynkin indices (4,0,0,0)</desc> <g transform="translate(1,1)"> <use xlink:href="#rect123"/> <use transform="translate(10,0)" xlink:href="#rect123"/> <use transform="translate(20,0)" xlink:href="#rect123"/> <use transform="translate(30,0)" xlink:href="#rect123"/> </g> </svg>\end{svg}}} - \underset{\color{brown}\frac{(n-1)n(n+1)(n+2)}{8}}{\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}}}+ \underset{\color{brown}\frac{(n+1)n(n-1)(n-2)}{8}}{\array{\begin{svg}<svg xmlns="http://www.w3.org/2000/svg" xmlns:xlink='http://www.w3.org/1999/xlink' width="1.5em" height="2em" viewBox="0 0 24 32"> <g transform="translate(1,1)"> <desc>Young Diagram with Dynkin indices (1,0,1,0)</desc> <use xlink:href="#rect123"/> <use transform="translate(0,10)" xlink:href="#rect123"/> <use transform="translate(0,20)" xlink:href="#rect123"/> <use transform="translate(10,0)" xlink:href="#rect123"/> </g> </svg>\end{svg}}} - \underset{\color{brown}\frac{n(n-1)(n-2)(n-3)}{4!}}{\array{\begin{svg}<svg xmlns="http://www.w3.org/2000/svg" xmlns:xlink='http://www.w3.org/1999/xlink' width=".75em" height="2.75em" viewBox="0 0 12 44"> <desc>Young Diagram with Dynkin indices (0,0,0,1)</desc> <g transform="translate(1,3)"> <use xlink:href="#rect123"/> <use transform="translate(0,10)" xlink:href="#rect123"/> <use transform="translate(0,20)" xlink:href="#rect123"/> <use transform="translate(0,30)" xlink:href="#rect123"/> </g> </svg>\end{svg}}} \end{aligned}

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

The relation ψ 4=ψ 2ψ 2\psi_4 = \psi_2\circ\psi_2 follows from

Sym 2( Young Diagram with Dynkin indices (2,0) ) = Young Diagram with Dynkin indices (4,0,0,0) + Young Diagram with Dynkin indices (0,2,0,0) n 2(n 21)12 2( Young Diagram with Dynkin indices (2,0) ) = Young Diagram with Dynkin indices (2,1,0,0) Sym 2( Young Diagram with Dynkin indices (0,1) ) = Young Diagram with Dynkin indices (0,0,0,1) + Young Diagram with Dynkin indices (0,2,0,0) 2( Young Diagram with Dynkin indices (0,1) ) = Young Diagram with Dynkin indices (1,0,1,0) \begin{aligned} \mathop{Sym}^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="2.75em" height=".75em" viewBox="0 0 44 12"> <desc>Young Diagram with Dynkin indices (4,0,0,0)</desc> <g transform="translate(1,1)"> <use xlink:href="#rect123"/> <use transform="translate(10,0)" xlink:href="#rect123"/> <use transform="translate(20,0)" xlink:href="#rect123"/> <use transform="translate(30,0)" xlink:href="#rect123"/> </g> </svg>\end{svg}} + \underset{\color{brown}\frac{n^2(n^2-1)}{12}}{\array{\begin{svg}<svg xmlns="http://www.w3.org/2000/svg" xmlns:xlink='http://www.w3.org/1999/xlink' width="1.5em" height="1.5em" viewBox="0 0 24 24"> <desc>Young Diagram with Dynkin indices (0,2,0,0)</desc> <g transform="translate(1,3)"> <use xlink:href="#rect123"/> <use transform="translate(10,0)" xlink:href="#rect123"/> <use transform="translate(0,10)" xlink:href="#rect123"/> <use transform="translate(10,10)" xlink:href="#rect123"/> </g> </svg>\end{svg}}} \\ \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}}\\ \mathop{Sym}^2\left(\array{\begin{svg}<svg xmlns="http://www.w3.org/2000/svg" xmlns:xlink='http://www.w3.org/1999/xlink' width=".75em" height="1.5em" viewBox="0 0 12 24"> <desc>Young Diagram with Dynkin indices (0,1)</desc> <g transform="translate(1,3)"> <use xlink:href="#rect123"/> <use transform="translate(0,10)" 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=".75em" height="2.75em" viewBox="0 0 12 44"> <desc>Young Diagram with Dynkin indices (0,0,0,1)</desc> <g transform="translate(1,3)"> <use xlink:href="#rect123"/> <use transform="translate(0,10)" xlink:href="#rect123"/> <use transform="translate(0,20)" xlink:href="#rect123"/> <use transform="translate(0,30)" xlink:href="#rect123"/> </g> </svg>\end{svg}} + \array{\begin{svg}<svg xmlns="http://www.w3.org/2000/svg" xmlns:xlink='http://www.w3.org/1999/xlink' width="1.5em" height="1.5em" viewBox="0 0 24 24"> <desc>Young Diagram with Dynkin indices (0,2,0,0)</desc> <g transform="translate(1,3)"> <use xlink:href="#rect123"/> <use transform="translate(10,0)" xlink:href="#rect123"/> <use transform="translate(0,10)" xlink:href="#rect123"/> <use transform="translate(10,10)" xlink:href="#rect123"/> </g> </svg>\end{svg}}\\ \wedge^2\left(\array{\begin{svg}<svg xmlns="http://www.w3.org/2000/svg" xmlns:xlink='http://www.w3.org/1999/xlink' width=".75em" height="1.5em" viewBox="0 0 12 24"> <desc>Young Diagram with Dynkin indices (0,1)</desc> <g transform="translate(1,3)"> <use xlink:href="#rect123"/> <use transform="translate(0,10)" 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="1.5em" height="2em" viewBox="0 0 24 32"> <desc>Young Diagram with Dynkin indices (1,0,1,0)</desc> <g transform="translate(1,1)"> <use xlink:href="#rect123"/> <use transform="translate(0,10)" xlink:href="#rect123"/> <use transform="translate(0,20)" xlink:href="#rect123"/> <use transform="translate(10,0)" xlink:href="#rect123"/> </g> </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=5k=5 case), it seems natural to conjecture that the kth Adams operation is the alternating sum of “hook representations” H s k(E)=s{ Young diagram with Dynkin indices (k-s-1,0,...,0,1) ks,rank(H s k(E))=(n+ks1)!k(ns1)!(ks1)!s! 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}<svg xmlns="http://www.w3.org/2000/svg" xmlns:xlink='http://www.w3.org/1999/xlink' width="3.25em" height="2.75em" viewBox="0 0 52 44"> <desc>Young diagram with Dynkin indices (k-s-1,0,...,0,1)</desc> <g transform="translate(1,1)"> <use xlink:href="#rect123"/> <use transform="translate(10,0)" xlink:href="#rect123"/> <use transform="translate(20,0)" xlink:href="#rect123"/> <use transform="translate(30,0)" xlink:href="#rect123"/> <use transform="translate(40,0)" xlink:href="#rect123"/> <use transform="translate(0,10)" xlink:href="#rect123"/> <use transform="translate(0,20)" xlink:href="#rect123"/> <use transform="translate(0,30)" xlink:href="#rect123"/> </g> </svg>\end{svg}}}}},\qquad \mathop{rank}(H^k_s(E)) = \frac{(n+k-s-1)!}{k(n-s-1)!(k-s-1)!s!} with kk boxes:

(1)ψ k(E)= s=0 k1(1) sH s k(E)\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 EE, and check that it behaves correctly for the direct sum with a line bundle. If we denote H 1 j(E)δ j,0H^j_{-1}(E)\equiv \delta_{j,0}, we can compactly write H s k(E+L)= j=1 ksH s1 kj(E)L j+ j=0 ks1H s kj(E)L j 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 s=0 k1(1) sH s k(E+L)=L k+ s=0 k1(1) sH s k(E) \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 s k(LE)=L kH s k(E) 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)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 kk-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 MM be the (k1)(k-1)-dimensional “Standard” representation of S kS_k (the quotient of the kk-dimensional “permutation” representation by the 1-dimensional trivial representation). The kk-hooks are the exterior powers of MM. In other words, the RHS of (1) corresponds to the element 1(M)= s(1) s s(M) \wedge_{-1}(M) = \sum_s (-1)^s \wedge^s(M) of the representation ring of S kS_k.

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

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6 Comments & 0 Trackbacks

Re: Adams Operations

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

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

for k=1,,nk=1,\dots,n and solve for ψ n\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

Re: Adams Operations

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)ψ k(E+1)=ψ k(E)+1, \psi_k(E+1) = \psi_k(E)+1,

where 11 is the trivial line bundle. But naive dimensional reduction yields three terms:

  1. all indices in EE
  2. one index in 11, k1k-1 indices in EE.
  3. all indices in the trivial bundle 11.

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

In other words: ψ k\psi_k is some expression in kk-block young diagrams, corresponding to representations of the symmetric group S kS_k. If we restrict to S k1S_{k-1}, this must vanish.

In Young diagram language, restricting to S k1S_{k-1} is the sum of all Young diagrams with one box removed. In particular, the kk-hook decomposes into two (k1)(k-1)-hooks, the kk-row into a (k1)(k-1)-row, and the kk-column into a (k1)(k-1)-column. (How did you draw those diagrams in itex?)

Now its easy to see that

(2)ψ k(E)=Sym k(E)+(stuff)=(k-row)+(stuff). \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 k1S_{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 ψ k(E+1)=ψ k(E)+1\psi_k(E+1) = \psi_k(E) + 1 where 11 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

2( Young Diagram with Dynkin indices (2,0) )= Young Diagram with Dynkin indices (2,1,0,0) \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}}

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

Re: Adams Operations

Now I’m curious. Why are you particularly interested in Adams operations?

Posted by: Andrew Stacey on June 4, 2009 8:09 AM | Permalink | Reply to this

Re: Adams Operations

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 λ\lambda-rings. You’re probably doing ‘orthodox’ stuff, since you’re talking about vector bundles. Borger is doing ‘unorthodox’ stuff: Frobenius lifts.

Posted by: John Baez on July 27, 2009 11:11 AM | Permalink | Reply to this

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