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(my question is also meaningful for complex K-theory, but since Kn(X) is always isomorphic to K-n(X), it's less interesting)


I start by recalling the analytic definition of KO-theory:

The following ingredients are needed:

Clifford algebras:
For n∈ℤ, the Clifford algebra Cliff(n) is the following ℤ/2-graded C*-algebra:
• $\langle e_1,\ldots, e_n\;|\; e_i \text{ is odd}, e_i^2=1, e_ie_j=-e_je_i, e_i^*=e_i\rangle$ if n ≥ 0.
• $\langle e_1,\ldots, e_{-n}\;|\; e_i \text{ is odd}, e_i^2=-1, e_ie_j=-e_je_i, e_i^*=-e_i\rangle$ if n ≤ 0.
Note: The above definition might seem a bit weird with its two cases n ≥ 0 and n ≤ 0, but actually it's quite ok: the map n $\mapsto$ Cliff(n) can be extended to a homomorphism from the associative group (ℤ,+) to the monoidal 2-category of ℤ/2-graded C${}^*$-algebras, bimodules, and bimodule maps.

Fredholm operators:
If H is a Hilbert space, then an operator F : HH is called Fredholm if boths its kernel and cokernel are finite dimensional.

The definition:

Let X be a topological space. A class in KOn(X) is represented by:

Option #1:
• A bundle of ℤ/2-graded real Hilbert spaces over X.
• Actions of Cliff(n) on the fibers of the above bundle.
• Fiberwise Fredholm operators that are odd, Cliff-linear, and skew-adjoint.
Option #2:
• A bundle of ℤ/2-graded real Hilbert spaces over X.
• Actions of Cliff(-n) on the fibers of the above bundle.
• Fiberwise Fredholm operators that are odd, Cliff-linear, and self-adjoint.

Dually, a class in KO-n(X) can be represented by:

Option #1:
• A bundle of ℤ/2-graded real Hilbert spaces over X.
• Actions of Cliff(n) on the fibers of the above bundle.
• Fiberwise Fredholm operators that are odd, Cliff-linear, and self-adjoint.
Option #2:
• A bundle of ℤ/2-graded real Hilbert spaces over X.
• Actions of Cliff(-n) on the fibers of the above bundle.
• Fiberwise Fredholm operators that are odd, Cliff-linear, and skew-adjoint.


If the bundles are finite dimensional, then the Fredholm operators can be taken to be zero. As a consequence, we get two maps

  • Bundles of finite dimensional Cliff(n)-modules over X${}\to KO^n(X)$

  • Bundles of finite dimensional Cliff(n)-modules over X${}\to KO^{-n}(X)$

In other words, a bundle of finite dimensional Cliff(n)-modules represents both an element in KOn(X) and in KO-n(X)!

Is there a homotopy-theoretical explanation of the above phenomenon?
Are there other cohomology theories E that have functors mapping to both En(X) and E-n(X)?
Maybe for E=TMF?

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5  
This question captures much of what I've never understood about Bott periodicity. I think of it this way: you have maps $f:K(Cliff(n))\rightleftarrows K(Cliff(n+1)): g$, where $f$ comes from tensoring up along a ring homomorphism $Cliff(n)\to Cliff(n+1)$, and $g$ from restricting along the same homomorphism. The fibers of the $f$'s give you the spaces in the $KO$-spectrum; the fibers of the $g$'s give you the same spaces in the opposite order! Weird. –  Charles Rezk Apr 22 '11 at 23:01
    
Am I correct that in the definition of $\operatorname{Cliff}(-n)$, you mean for the sequence to start at $e_{-1}$? –  Theo Johnson-Freyd Apr 23 '11 at 0:12
3  
@Theo: The algebras under consideration are real, not complex. –  Dmitri Pavlov Apr 23 '11 at 1:39
3  
Maybe Kasparov theory helps to tidy the confusion a bit. Option #2 (in the first grey box) amounts to an element in $KK(R; C(X) \otimes Cl (-n))$, while option #1 (also in the first box) amounts to an element in $KK(Cl(n);C(X))$. To get this from option #1, multiply the Fredholm with the grading involution and obtain a self-adjoint one that anticommutes with $Cl(n)$. Both groups are isomorphic (by ''abstract Bott periodicity''). –  Johannes Ebert Apr 23 '11 at 11:30
1  
@Theo: In the definition of Cliff(n) for negative $n$, the sequence of generators starts at $e_1$, and goes until $e_{-n}$ (recall that -n is positive). –  André Henriques Apr 23 '11 at 19:17

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