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Let $\mathbb T^2$ be the 2-Torus and let $X$ be a topological space.

Is there any way of computing $[\mathbb T^2,X]$, the set of homotopy class of continuous maps $\mathbb T^2\to X$ if I know, for instance, the homotopy groups of $X$?

Actually, I am interested in the case $X=\mathbb{CP^\infty}$. I would like to classify $\mathbb T^1$-principal bundles over $\mathbb T^2$ (in fact $\mathbb T^2$-principal bundles, but this follows easily.)

Rasmus
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    As I said on your other question, $\mathbb{CP}^{\infty}$ is a $K(\mathbb{Z}, 2)$, so there is a natural bijection from $[\mathbb{T}^2, \mathbb{CP}^{\infty}]$ to $H^2(\mathbb{T}^2; \mathbb{Z}) \cong \mathbb{Z}$. – Qiaochu Yuan May 02 '11 at 19:22
  • @Qaiochu: Of course. I should have realised this myself. Thank you! – Rasmus May 02 '11 at 19:40
  • I would still be interested in an answer for general $X$, even though I don't expect a simple answer. – Rasmus May 02 '11 at 19:44

3 Answers3

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(Disclaimer. The previous version of this answer contained serious mistake: it didn't take into account non-triviality of the action of $\pi_1$ on $\pi_2$.)

Claim. The set $[\mathbb T^2,X]_*$ (of pointed maps) can be identified with the set $$\{(a,b)\in\pi_1(X)^2|ab=ba\}\times\pi_2(X)/\langle t-t^a,t-t^b\mid t\in\pi_2\rangle,$$ where $(-)^\gamma$ denotes the action of $\pi_1$ on $\pi_n$.

And if $\pi_1$ acts trivially on $\pi_2$, $[\mathbb T^2,X]_\ast\approx\{(a,b)\in\pi_1(X)^2|ab=ba\}\times\pi_2(X)$. In particular, if $\pi_1(X)=0$, $[\mathbb T^2,X]_\ast=[\mathbb T^2,X]\cong\pi_2(X)=H_2(X)$.

Proof (sketch). Indeed, parallel and meridian of $\mathbb T^2$ maps to a pair of elements of $\pi_1(X)$ and the 2-cell of the torus maps to a null-homotopy of $aba^{-1}b^{-1}$, but homotopies between trivial loop and some other null-homotopic loop can be identified (non-canonically!) with $\pi_2(X)$.

Let's see if this element of $\pi_2$ is well-defined. If we move $a$ by some (pointed) homotopy $t$ (that can be again identified with an element of $\pi_2$), we get $s'=s+t-t^b$ (in particular, if either both $a$ and $b$ are trivial or $\pi_1(X)$ acts trivially on $\pi_2(X)$, the element $s$ is well-defined).

/* This kind of elementary obstruction theory (cf.) can be applied, I believe, to any 2-dimensional CW-complex $S$ giving (in the case $\pi_1(X)$ acts trivially on $\pi_2(X)$) $[S,X]\approx H^1(S;\pi_1(X))\times H^2(S;\pi_2(X))$. But in higher dimensions the situation becomes more complicated. */

So, for example, $$[\mathbb T^2,\mathbb CP^\infty]\cong\pi_2(\mathbb CP^\infty)\cong\mathbb Z.$$ But, say, $$[\mathbb T^2,\mathbb R P^2]_\ast\cong(\mathbb Z/2)^2\times\mathbb Z/{\sim}\cong\{(0,0)\}\times \mathbb Z\sqcup\{(0,1),(1,0),(1,1)\}\times\mathbb Z/2$$ (if the element of $\pi_1^2$ is nontrivial, the element of $\pi_2$ is defined only mod 2; and in non-pointed case all $\mathbb Z$ become $\mathbb N$; ref).

Community
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Grigory M
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  • That's very interesting, thanks. (I fixed a tex-typo.) – Rasmus May 02 '11 at 20:21
  • That's very interesting. Can you provide some details about this? And I'm curious whether this kind of argument can be applied to other CW-complexes for example surfaces with higher genus, and other well-known spaces. – Honglu Nov 15 '11 at 03:37
  • @Grigory: I'm a little confused by this answer. When I apply this argument to $D^2$ rather than $T^2$ (thinking of $D^2$ as a loop and then a null-homotopy of this loop) I don't seem to get the right thing. – Qiaochu Yuan Jul 15 '14 at 06:37
  • @Qiaochu Let's see. In the same spirit for $D^2$ we have 'a' trivial element of $\pi_1$ and an element of $\pi_2$ up to some equivalence relation; and in the part 'Let's see if this element of $\pi_2$ is well-defined' we have $s'=s+t$ — i.e. equivalence relation identifies all elements of $\pi_2$. That's the correct answer. – Grigory M Jul 15 '14 at 09:51
  • @Grigory: so $t^b$ vanishes here? But if $b$ is trivial we should get $t^b = t$, not $t^b = 0$, right? Probably I'm misunderstanding something. – Qiaochu Yuan Jul 15 '14 at 18:20
  • @Qiaochu The answer for $D^2$ is different because the coboundary map is different. For $[S,X]$ (and let's consider the case $\pi_1(X)=0$ for simplicity) we're talking about $H^2(S;\pi_2(X))$. – Grigory M Jul 15 '14 at 18:30
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    @Qiaochu ...i.e. $\pi_2(X)/\langle t-t^a,t-t^b\mid t\in\pi_2\rangle$ is actually $H^2(\mathbb T^2;\pi_2(X))$, where $\pi_2(X)$ is the local system s.t. generators of $\pi_1(\mathbb T)$ act as $t\mapsto t^a$ and $t\mapsto t^b$. – Grigory M Jul 15 '14 at 18:33
  • (I wish I knew a reference where all this is explained more clearly...) – Grigory M Jul 15 '14 at 18:35
  • @Grigory: Oh, I see. And in the case of $D^2$ there isn't an interesting local system in play. Thanks for the clarification! I too wish there was a reference for this. – Qiaochu Yuan Jul 15 '14 at 18:36
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This is a good chance to advertise the paper

Ellis, G.J. Homotopy classification the J. H. C. Whitehead way. Exposition. Math. 6(2) (1988) 97-110.

Graham Ellis is referring to Whitehead's paper "Combinatorial Homotopy II", not so well read as "Combinatorial Homotopy I".

He writes:" Almost 40 years ago J.H.C. Whitehead showed in \cite{W49:CHII} that, for connected $CW$-complexes $X, Y$ with dim $X \le n$ and $\pi_i Y = 0$ for $2\le i \le \ n - 1$, the homotopy classification of maps $X \to Y$ can be reduced to a purely algebraic problem of classifying, up to an appropriate notion of homotopy, the $\pi_1$-equivariant chain homomorphisms $C_* \widetilde{X} \to C_* \widetilde{Y}$ between the cellular chain complexes of the universal covers. The classification of homotopy equivalences $Y \simeq Y$ can similarly be reduced to a purely algebraic problem. Moreover, the algebra of the cellular chains of the universal covers closely reflects the topology, and provides pleasant and interesting exercises.

"These results ought to be a standard piece of elementary algebraic topology. Yet, perhaps because of the somewhat esoteric exposition given in \cite{W49:CHII}, and perhaps because of a lack of worked examples, they have remained largely ignored. The purpose of the present paper is to rectify this situation."

Ronnie Brown
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If you want to calculate $[\mathbb T^2,\mathbb CP^\infty]$, perhaps, it's easier to use the classification of maps to $\mathbb CP^\infty$ instead: $[X,\mathbb CP^\infty]=H^2(X)$; so $[\mathbb T^2,\mathbb CP^\infty]=H^2(\mathbb T^2)=\mathbb Z$.

Grigory M
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