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After Bott periodicity is proved, one still has to compute the stable values. For the unitary group $U$, this is easy since you can get away with just $\pi_0$ and $\pi_1$. However, I'm having trouble doing this for the orthogonal group. It's easy to work out $\pi_i(O)$ for, say, $0 \leq i \leq 2$, but already for $i=3$ it takes some work. Can anyone either give me a reference or proof for the computation of $\pi_i(O)$ for $0 \leq i \leq 7$?

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  • $\begingroup$ Thanks to everyone. I accepted Andre's answer, but all of them were great and much appreciated. $\endgroup$ – Linda Jan 7 '15 at 20:18
  • $\begingroup$ There is another way to compute it: Milnor or Stong found cohomology of bo as an algebra over Steendrod algebra, so when we take it as input of Adams spectral sequence, we hope we get something computable. In this case it is. We can solve the whole spectral sequence and see the Bott periodicity there. The detailed computation is in Ravenel's Green Book page 64-66 $\endgroup$ – Mingcong Zeng Jan 11 '15 at 4:48
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Bott's result is best formulated by saying that the sequence of spaces $$ \mathbb Z\times BO,\quad O,\quad O/U,\quad U/Sp,\quad \mathbb Z\times BSp,\quad Sp,\quad Sp/U,\quad U/O $$ are related by the property that each one is the loop space of the previous one (mod 8).

From that, you get for example that $\pi_4(O)=\pi_3(O/U)=\pi_2(U/Sp)=\pi_1(Z\times BSp)=\pi_0(Sp)=0$.

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You might like to look at my thesis, where a lot of this kind of thing is recalled in a consolidated way: http://neil-strickland.staff.shef.ac.uk/research/thesis.pdf

Some key points:

  • The homotopy groups of the space $O$ are the same as the homotopy groups of the spectrum $KO$, shifted by one. It is easier to think about $KO$, because it has a ring structure.
  • The best approach is to admit that you actually know the answer already, and prove that it is correct, rather than pretending that you need to find the answer from scratch.
  • To produce the generators and relations, you can use Atiyah's theory of Clifford modules.
  • To show that there are no further generators and relations, we need some exact sequences. The main point is that there is a cofibration $\Sigma KO\xrightarrow{\eta}KO\xrightarrow{}KU\xrightarrow{}\Sigma^2KO$.
  • Many parts of the story are most efficiently organised using the ideas in Atiyah's paper "K-Theory and Reality".
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This is a long comment on André's answer. Here is a way to organize the spaces that appear in it using the categories of $\mathbb{Z}_2$-graded modules over Clifford algebras.

Let $\text{Cliff}(p, q)$ denote the Clifford algebra obtained by adjoining $p$ anticommuting square roots of $1$ and $q$ anticommuting square roots of $-1$ to $\mathbb{R}$, and let $\text{Mod}(n)$ denote the category of $\mathbb{Z}_2$-graded modules over $\text{Cliff}(p, q)$ for any $p, q$ such that $p - q \equiv n \bmod 8$ (this is part of the Clifford-algebraic statement of Bott periodicity). Then:

  • $\text{Mod}(0) \cong \text{Vect}_{\mathbb{R}} \times \text{Vect}_{\mathbb{R}}$.
  • $\text{Mod}(\pm 1) \cong \text{Vect}_{\mathbb{R}}$.
  • $\text{Mod}(\pm 2) \cong \text{Vect}_{\mathbb{C}}$.
  • $\text{Mod}(\pm 3) \cong \text{Vect}_{\mathbb{H}}$.
  • $\text{Mod}(\pm 4) \cong \text{Vect}_{\mathbb{H}} \times \text{Vect}_{\mathbb{H}}$.

Each of these module categories is in particular symmetric monoidal under direct sum and so presents an infinite loop space, which I'll label $K(n)$. These infinite loop spaces are the following:

  • $K(0) \cong KO \times KO$.
  • $K(\pm 1) \cong KO$.
  • $K(\pm 2) \cong KU$.
  • $K(\pm 3) \cong KSp$.
  • $K(\pm 4) \cong KSp \times KSp$.

These categories and these infinite loop spaces can be organized into a clock in the obvious way. There are natural pairs of adjoint functors given by induction and restriction between any consecutive categories in this clock, and these induce maps on the corresponding infinite loop spaces. The following statement implies Bott periodicity in the strong form that André states it (which is the form in which Bott originally proved it).

Theorem: The homotopy fiber of the natural map $K(n) \to K(n-1)$ is also the homotopy cofiber of the natural map $K(n+1) \to K(n)$.

Corollary: The homotopy fiber of the natural map $K(n) \to K(n-1)$ and the homotopy cofiber of the natural map $K(n+1) \to K(n)$ are both the $n$-fold loop space $\Omega^n KO$, and $\Omega^8 KO \cong KO$.

On the other hand, you should be able to convince yourself at least heuristically that the homotopy fiber of the natural map $KO \times KO \to KO$ is $KO$, that the homotopy fiber of the natural map $KU \to KO$ is $O/U$, and so forth. In other words, you can rewrite that sequence of spaces heuristically as

$$\frac{KO \times KO}{KO}, \frac{KO}{KO \times KO}, \frac{KU}{KO}, \frac{KSp}{KU}, \frac{KSp \times KSp}{KSp}, \frac{KSp}{KSp \times KSp}, \frac{KU}{KSp}, \frac{KO}{KU}.$$

where $\frac{X}{Y}$ is heuristic notation for the homotopy fiber of a map $X \to Y$, and in particular $\frac{1}{BG}$ is $G$, so that $\frac{KU}{KO}$ becomes $O/U$ and so forth.

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    $\begingroup$ If memory serves, that's the approach taken in Milnor's "Morse Theory," which gives a very concrete proof of Bott periodicity. $\endgroup$ – anomaly Jan 7 '15 at 6:19
  • $\begingroup$ Hmm, I think when I wrote $KO$ above I should've written $ko$, etc. $\endgroup$ – Qiaochu Yuan Mar 14 '15 at 5:08
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A nice way to do this calculation can be extracted from the papers

[ABS] Atiyah, Bott, Shapiro: ''Clifford modules''

[AtKR] Atiyah: ''K-Theory and Reality''.

Let $A_n$ be the group of $Cl^n$-modules, modulo $Cl^{n+1}$-modules, as introduced by Atiyah, Bott, Shapiro. The sum of these is a graded ring, and the main result of [ABS] is that the homomorphism $$abs:A_{\ast} \to KO^{-\ast}(\ast)$$ they construct is a ring isomorphism. That $A_{\ast} = Z [\eta, \lambda, \beta] /(2 \eta, \eta^3, \lambda \eta, \lambda^4- 2 \beta)$ is done using linear algebra in [ABS]. The original proof in [ABS] used the known structure of $KO^{-\ast}$. In [AtKR], a simple proof of real periodicity is given, and using ideas from [AtKR], one can give a proof that $abs$ is an iso without using that knowledge, and thereby compute the $KO$-groups.

The ingredients one needs are

  1. $8$-periodicity, or more precisely, that multiplication by $abs (\beta )\in KO^{-8}$ is an isomorphism.
  2. The knowledge of the complex $K$-groups.
  3. That the complexification $KO^0 \to K^0$ is an isomorphism.
  4. That $KO^{-1} =Z/2$, generated by $abs(\eta)$. This is the easy fact that $O(n)$ has two components, in disguise.
  5. The long exact sequence (proven in [AtKR], § 3)

$$\ldots KO^{1-q} \stackrel{\eta}{\to} KO^{-q} \to K^{-q} \to KO^{2-q} \ldots,$$

the map to complex $K$-theory is complexification, and we write $\eta:= abs (\eta)$ and use the same letter for the multiplication by $\eta$.

First look at the piece $$ K^{-3} \to KO^{-1} \to KO^{-2} \to K^{-2} \to KO^0 \to KO^{-1} \to K^{-1} \to KO^1\to KO^0 \to K^0$$

As $KO^0 \to K^0$ is an iso and $K^{-1}=0$, you get that $KO^1 =0$. Since $KO^{-1}=Z/2$, the map $Z= K^{-2} \to KO^{0} = Z$ must be multiplication by $\pm 2$ and hence be injective. Therefore, multiplication by $\eta$ is surjective $KO^{-1} \to KO^{-2}$. Since $K^{-3}=0$, it is also injective and you get $KO^{-2} = Z/2$, the element $\eta^2 $ is nonzero. Continue with

$$ 0= K^{-5} \to KO^{-3} \to KO^{-4} \to K^{-4} \to KO^{-2}\to KO^{-3} \to K^{-3}=0$$

Hence $\eta: KO^{-2} \to KO^{-3}$ is surjective, but as the only nonzero element of $KO^{-2}$ is $\eta^2$ and since $\eta^3 =0$ (this follows from the corresponding relation in the algebraic model $A_{\ast}$), $KO^{-2} \to KO^{-3}$ is also null. Thus $KO^{-3}=0$. It follows that $KO^{-4}=Z$, and that $KO^{-4} \to K^{-4}$ is multiplication by $\pm 2$. The last two portions of the long exact sequence are

$$ 0=K^{-7} \to KO^{-5} \to KO^{-6} \to K^{-6} \to KO^{-4} \to KO^{-5} \to K^{-5}=0 $$

and

$$ KO^{-8} \stackrel{\cong}{\to} K^{-8} \to KO^{-6} \to KO^{-7} \to K^{-7}=0. $$

We have seen that $KO^{-7}=KO^1 =0$, thus $KO^{-6}=0$ by the second sequence. This shows that $KO^{-5}=0$ (use first sequence) and completes the argument.

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