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Let $T^n$ denote $n$-torus defined as:

$$ T^n = \underbrace{S^1\times S^1 \times \dots \times S^1}_n $$

The question is to compute $H_k(T^n)$ without using Kuneth formula.


So far I've noticed that $T^n$ can be decomposed into two homotopic parts parts with intersection homotopic to two copies of the parts s.t.

$$ T^n = A \cup B \\ B \simeq A \simeq [0,1] \times T^{n-1} \simeq T^{n-1} \\ A \cap B \simeq T^{n-1} \coprod T^{n-1}$$

Knowing that homology of disjoint union is direct product of homologies, this gives following exact sequence:

$$ \dots \rightarrow H_{k+1}(T^{n+1}) \rightarrow H_k(T^n)^2 \rightarrow H_k(T^n)^2 \rightarrow H_k( T^{n+1} ) \rightarrow \dots $$

I know additionally homology of $T^2$. Is this sufficient information to finish this?

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    $\begingroup$ $(T^n)^2$ is not the same as $T^n \coprod T^n$. Also, just knowing the terms in the Mayer-Vietoris sequence is rarely enough (unless there's a lot of zero terms). You must also know the homomorphisms. $\endgroup$
    – Lee Mosher
    May 14, 2019 at 14:10
  • $\begingroup$ The square there was to denote $H_k(T^n) \oplus H_k(T^n)$. $\endgroup$
    – Radost
    May 14, 2019 at 16:57
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    $\begingroup$ Alright. But, the rest of my comment stands: you need formulas for all the homomorphisms represented by horizontal arrows. Without knowing that, the problem is essentially impossible, except in very extreme cases such as when a lot of terms of the Mayer-Vietoris sequence are known to be trivial groups. $\endgroup$
    – Lee Mosher
    May 14, 2019 at 18:05
  • $\begingroup$ I'm guessing it's not the case and any induction fails? $\endgroup$
    – Radost
    May 14, 2019 at 19:11
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    $\begingroup$ On the contrary, it should be possible to write down formulas for those homomorphisms. $\endgroup$
    – Lee Mosher
    May 14, 2019 at 19:12

1 Answer 1

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Yeah, this works! In fact, not yet knowing things like the Künneth formula, this was the solution I posed to myself to explain the formula $H_k(T_n)\cong\Bbb Z^{\binom{n}{k}}$ in as elementary a fashion as I could. In fact, there is a general phenomenon we could appeal to here.

Let $X$ be any non-empty space; let $k$ be any integer greater than $0$.

There exists an exact sequence: $$0\to H_k(X)\to H_k(X\times S^1)\to H_{k-1}(X)\to0$$

Proof - following your chosen application of Mayer-Vietoris. Let $A$ and $B$ be proper open arcs of $S^1$, chosen so that they intersect in two disjoint arcs. It’s easy to explicitly describe such things. Then observe that $X\times A$ and $X\times B$’s interiors cover $X\times S^1$, each is homotopy equivalent to $X$ and furthermore their intersection $X\times(A\cap B)$ is homotopy equivalent to $X\sqcup X$. According to the Mayer-Vietoris theorem, there is an exact sequence: $$\cdots\to H_k(X\times(A\cap B))\to H_k(X\times A)\oplus H_k(X\times B)\to H_k(X\times S^1)\to H_{k-1}(X\times(A\cap B))\to H_{k-1}(X\times A)\oplus H_{k-1}(X\times B)\to\cdots$$And we have the privilege of knowing full descriptions of all maps involved, more or less. In particular the first and last maps take cycles $x$ of the intersection to the pair $(x,-x)$. Under choices of isomorphism $H_k(X\times A)\cong H_k(X)$ etc. (e.g. strong deformation retractions), since they take generators to generators and relabelling of generators is just another isomorphism, not affecting exactness, it’s clear that there is an equivalent model of this sequence: $$\cdots\to H_k(X\sqcup X)\cong H_k(X)\oplus H_k(X)\to H_k(X)\oplus H_k(X)\overset{\alpha}{\longrightarrow}H_k(X\times S^1)\overset{\beta}{\longrightarrow}H_{k-1}(X)\oplus H_{k-1}(X)\to H_{k-1}(X)\oplus H_{k-1}(X)\to\cdots$$The first and last maps act algebraically by taking, on each component, $a\mapsto(a,-a)$ - so their full description is $(a,b)\mapsto(a+b,-a-b)$.

$\alpha$ has kernel equal to the image $\langle a+b,-(a+b)\rangle$ which is just to say there is an injection $H_k(X)^2/\langle a+b,-(a+b)\rangle\to H_k(X\times S^1)$; by, say, the first isomorphism theorem, it’s clear the left hand side is an isomorph of $H_k(X)$. Similarly, since the image of $\beta$ is just the kernel of the same map (but on dimension $(k-1)$) and this kernel $\langle a+b,-(a+b)\rangle$ is clearly also an isomorph of $H_{k-1}(X)$ we find a surjection $H_k(X\times S^1)\to H_{k-1}(X)$. Since $\alpha$ and $\beta$ are exact at $H_k(X\times S^1)$, the induced diagram $H_k(X)\to H_k(X\times S^1)\to H_{k-1}(X)$ is also exact! This concludes the proof.

That was verbose so that everything is crystal clear (hopefully); the actual computation here was very easy and used more elementary tools. So the good news is that, if $H_{k-1}(X)$ is free, there must exist (splitting lemma) an isomorphism $H_k(X\times S^1)\cong H_k(X)\oplus H_{k-1}(X)$.

Apply this now to the tori $X=T_n=(S^1)^n$ for $n\ge1$. Since each of these is path-connected, we only need to keep track of their homology in degree $k\ge1$. Assume for the sake of induction we have $H_k(T_n)\cong\Bbb Z^{\binom{n}{k}}$ for $k\ge1$ and some $n$. Of course, this is true for $n=1$ by an elementary calculation.

The above result (inductive hypothesis says the group $H_{k-1}(X)$ is free) gives $H_k(T_{n+1})\cong\Bbb Z^{\binom{n}{k}}\oplus\Bbb Z^{\binom{n}{k-1}}\cong\Bbb Z^{\binom{n+1}{k}}$ for $k\ge1$, as desired. The induction turns out to be very easy!

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