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I find in my books it is given by Bott periodicity, but this is not direct and Bott periodicity is not easy. Is there an easy and direct way to define $K^n(X)$, like $K^{-n}(X)$? I just start to learn this stuff...

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Do they do K-theory on Mars, where you seem to be from? –  Alex B. Oct 19 '10 at 7:47
    
Can't you just suspend $X$ a few times and take maps to BO? –  S. Carnahan Oct 19 '10 at 7:48
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@Scott Carnahan: that produces negative K-theory. –  Johannes Ebert Oct 19 '10 at 8:17
    
Thanks, my mistake. –  S. Carnahan Oct 19 '10 at 12:30
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3 Answers 3

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There is a definition of $K^n$ for positive $n$, without Bott periodicity. This approach goes back to Karoubi, and you can find it in his book "K-theory". The definition (for both, positive and negative $n$) uses Clifford algebras, but no Bott periodicity (Karoubi uses his new, more algebraic and direct definition to prove Bott periodicity). On the other hand, without Bott periodicity, topological $K$-theory is not very interesting. Also, the identification of Karoubi's negative $K$-theory with the ordinary theory is not a triviality - this is more or less equivalent to Bott periodicity.

Another way to phrase the problem: To define $K^{-n}(X)$, you can take the n-fold suspension of $X$ or the n-fold loop space of $Z \times BU$. If you wish to define $K^n (X)$, you need a space $Y_n$ whose $n$-fold loop space is $Z \times BU$. So what you need to know is that $Z \times BU$ is what topologists call an "infinite loop space", see Adams nice book with the same title. This is what Bott periodicity does, and it is absolutely crucial to define $K$-theory as a USEFUL cohomology theory.

As pointed in other answers to this question, there is another way to construct K-theory as a cohomology theory - the "infinite loop space machines". You can read in Adams book "Infinite loop spaces" about it - this is one of the most pleasant math books I know, not least because he leaves out all the technicalities. If you go into the details of the infinite loop space machines (May or Segal), it becomes WAY more technical than most the proofs of Bott periodicity, at least if you take the details seriously - it takes May about 80 pages to write everything down. Segals approach looks substantially easier, but that might be due to his very condensed writing.

Using this theory of infinite loop spaces, you get a cohomology theory k^n. This theory is closely related to, but by no means equal to the ordinary K-theory (it is connective, as is any spectrum coming from infinite loop space machines).

If you want to study K-theory seriously, there is absolutely no way around the Bott periodicity theorem: I am not aware of any application of K-theory that can be done without it. The reason is that you cannot compute anything without Bott periodicity. This starts with the first computations (spheres and projective spaces) and goes on to applications like Hopf-invariant one problem, vector fields on spheres, Adams work on Im (J), the Adams conjecture a la Becker-Gottlieb ..., not to mention Index theory.

Last but not least, there are several proofs of B.P. which I found very pleasant to read, for example Atiyah's "Bott periodicity and elliptic operators".

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Without Bott periodicity, any clue to BU can be delooped infinity times? –  Ma Ming Oct 19 '10 at 12:32
    
Aha! U(n) is homotopy-Abelian in U(2n) (See IOAN JAMES AND EMERY THOMAS "Which Lie Groups are Homotopy-Abelian"), therefore we may say U(\infty) is a homotopy-Abelian group (E_\infty space??). On the other hand, we know that Abelian groups are infinite loop space. Any chance to make this intuition be a concrete proof? –  Ma Ming Oct 19 '10 at 13:07
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I'd disagree with the claim "without Bott periodicity, K-theory is not very interesting". There is a cohomology theory called "connective K-theory" $k^*(X)$, obtained merely using the fact that $Z\times BU$ is an $E_\infty$-space, as in Justin's answer. The groups $k^q(X)$ vanish when $X$ is a space and $q>0$. You can obtain $K^*(X)$ from $k^*(X)$ by inverting the "Bott element" $b\in k^{-2}(*)$, but $K^*$ does not determine $k^*$. You shouldn't need Bott periodicity to carry out these constructions (though you'll certainly need periodicity to figure out what $k^*$ and $K^*$ look like.) –  Charles Rezk Oct 19 '10 at 18:31
    
"though you'll certainly need periodicity to figure out what k ∗ and K ∗ look like" - that is what I mean –  Johannes Ebert Oct 19 '10 at 18:40
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@Charles Rezk: Maybe that would open an expert's thread, but do you know of written account where the infinite-loop space viewpoint is explicitly combined with the Bott-periodicity? As far as I can see, this amounts to writing down all spaces of the Bott spectrum as $E_\infty$- or $\Gamma$-spaces, plus the Bott maps as maps preserving this structure. –  Johannes Ebert Oct 19 '10 at 18:45
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One way to do this is with infinite loop space machines, such as the one constructed by J.P. May in "The Geometry of Iterated Loop Spaces".

The main argument is that there is a contractible operad L such that BU is an L-algebra. The operad L is the complex linear isometries operad. Then, May's method proves that BU is an infinite loop space.

The argument goes roughly like this: Let $C_n$ be the little n-cubes operad, then using May's product trick, there are maps of operads: $L \longleftarrow L\times C_n \longrightarrow C_n$, the second of which is a weak equivalence.

Now, BU is also an $L\times C_n$ algebra. May shows that if $X$ is a nice space, then the free algebra $L\times C_n (X)$ is weakly equivalent to $\Omega^n\Sigma^n X$. Now, he applies a simplicial bar construction to obtain weak equivalences:

$BU \longleftarrow B(L\times C_n, L\times C_n, BU) \longrightarrow B(\Omega^n\Sigma^n, L\times C_n, BU) \to \Omega^n B(\Sigma^n, L\times C_n, BU)$

which is the required n-fold delooping. There are many details and a consistency check left to do, but this is the basic idea.

At this point you should be able to define $K^n(X) = [X, B(\Sigma^n, L\times C_n, BU)]$.

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I want to tag along on Johannes' answer, since I think it's helpful to see an explicit construction of the positive K-groups in terms of bundles of Clifford modules, due to Atiyah, Bott and Shapiro in Clifford Modules:

This construction works equally well for KU and KO, by taking complexified or real Clifford algebras, and I'll leave this out of the notation.

First some notation:

  • Let $\mathbb{R}^{p,q}$ denote $\mathbb{R}^{p+q}$ equipped with the quadratic form $-x_1^2-\ldots-x_q^2+x_{q+1}^2+\ldots+x_{q+p}^2$.
  • Let $C^{p,q}$ denote the associated Clifford algebra, and $M^{p,q}(X)$ denote the Grothendieck group of graded bundles of modules for $X\times C^{p,q}$.
  • Let $M^{p,q+1}(X)\rightarrow M^{p,q}(X)$ denote restriction of scalars associated to the embedding $\mathbb{R}^{p,q}\hookrightarrow \mathbb{R}^{p,q+1}$.

ABS construct a homomorphism $M^{p,q}(X)\rightarrow K(X\times B^{p+q}, X\times S^{p+q-1})$ which factors through $A^{p,q}(X):=M^{p,q}(X)/M^{p,q+1}(X)$. They use Bott periodicity to show that this is an isomorphism when $X$ is a point and $q=0$, and Karoubi essentially extends this for any $X$ under the identification $K^{q-p}(X)=K(X\times B^{p+q}, X\times S^{p+q-1})$ (disclaimer: Karoubi rephrases the construction in an algebraic formalism that I haven't completely translated into the ABS point of view. If anyone has, and can either confirm that it's the same or explain the difference, that would be really helpful).

The homomorphism is as follows:

Given a bundle of $X\times C^{p,q}$-modules $E=E^0\oplus E^1$, ABS observe that the Clifford action gives a bundle map $cl:E^0\rightarrow E^1$ on $X\times B^{p+q}$ which is an isomporphism on $X\times S^{p+q-1}$. This gives a class $[E^0,E^1,cl]\in K(X\times B^{p+q}, X\times S^{p+q-1})$.

If $E$ is in the image of $M^{p,q+1}(X)$, then the map $cl$ extends to an isomorphism on all of $X\times B^{p+q}$ (identify $B^{p+q}$ with the upper hemisphere of $S^{p+q}$), and $E$ is mapped to $0$, so the map indeed factors through $A^{p,q}(X)$.

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When you say “essentially extends this for any X”, do you mean that M^{p,q}(X) is isomorphic to K^{q−p}(X) for all p, q, and X? When you say “graded bundles of modules”, do you mean locally trivial finite-dimensional vector bundles equipped with an action of C^{p,q}? –  Dmitri Pavlov Nov 18 '12 at 23:17
    
I said "essentially" because I hadn't (and haven't) had a chance to thoroughly translate Karoubi's formalism into ABS. I understand the ABS map to be an isomorphism for all p and q, and nice X (probably finite CW complexes), but I haven't worked through it to the point where I can guarantee there's not a subtlety I'm missing. By a graded bundle of modules, I mean a Z/2-graded vector bundle (locally trivial, finite dimensional) equipped with an action of C^{p,q} respecting the grading and the fibers. –  Jesse Wolfson Nov 28 '12 at 23:46
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