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I'm reading a lecture notes on actions of $\mathbb{Z}^{p}$ on the torus but I think that are not selfcontained. I need to know the definition of the "induced action in homology" of a given action.

Let $p,q\in\mathbb{N}$ and $T=\mathbb{R}^{q}/\mathbb{Z}^{q}$ the $q-$dimensional torus. A linear action of the aditive group $\mathbb{Z}^{p}$ is any homomorphism of groups $\phi:\mathbb{Z}^{p}\rightarrow End(T^{q})$. Then, $\phi$ induces a linear action $\phi_{*}$ of $\mathbb{Z}^{p}$ on the first homology group $H_{1}(T^{q},\mathbb{Z})$. My question is

Which is the definition of $\phi_{*}$?

I will appreciate any reference on this particular topic.

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    $\begingroup$ any map of spaces induces a map on homology $\endgroup$ Dec 27, 2021 at 21:49
  • $\begingroup$ Do you have a link (or title & author) for the mentioned lecture notes? $\endgroup$
    – Alp Uzman
    Dec 28, 2021 at 4:41

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Roughly speaking, any element $\gamma$ of $H_1(\mathbb{T}^q;\mathbb{Z})$ can be thought of as a continuous loop in $\mathbb{T}^q$ (more accurately elements of $H_1$ are homotopy classes of such loops). In particular any such $\gamma$ can be written in the form

$$\gamma=\sum_{i=1}^q z_i \gamma_i,$$

where $z_i$'s are integers and for $1\leq i\leq q$, $\gamma_i: [0,1]\to \mathbb{T}^q$ is defined by $\gamma_i(t)=(0,\cdots,0,t,0,\cdots,0)$, where $t$ is in the $i$-th coordinate.

(See e.g. What does the element of homology groups mean? for a more accurate description.)

Example 1: For $q=2$, thinking of $\mathbb{T}^2$ as a square with opposite sides identified, $\gamma_1$ is the horizontal loop, $\gamma_2$ is the vertical loop, $\gamma_1+\gamma_2$ is the diagonal loop from the bottom left corner to top right corner and $-\gamma_1+\gamma_2$ is the diagonal loop from the bottom right corner to the top left corner.

After fixing the coordinate loops $\gamma_1,\cdots, \gamma_q$ since the integers $z_1,...,z_q$ completely determine $\gamma$ we have an isomorphism of groups:

$$H_1(\mathbb{T}^q;\mathbb{Z})\stackrel{\cong}{\to} \mathbb{Z}^q,\,\, \sum_{i=1}^q z_i \gamma_i \mapsto (z_1,\cdots ,z_q).$$

Next, if $f:\mathbb{T}^q\to \mathbb{T}^q$ is a continuous map (it does not need to be an endomorphism) and if $\gamma$ is a continuous loop in $\mathbb{T}^q$, then $f\circ \gamma$ is also a continuous loop in $\mathbb{T}^q$. Thus (as $f$ also preserves homotopies) there is an induced map $f_\ast: H_1(\mathbb{T}^q;\mathbb{Z})\to H_1(\mathbb{T}^q;\mathbb{Z})$.

Example 2: In a recent answer (https://math.stackexchange.com/a/4320757/169085) I've established that any Lie group endomorphism of $\mathbb{T}^q$ is given by a $q\times q$ matrix with integer entries. Fix $q=2$ and consider the endomorphism $f:\mathbb{T}^2\to\mathbb{T}^2, (x,y)\mapsto (2x,-3y)$. To see what the induced map $f_\ast$ does on homology, consider the coordinate loops:

$$f_\ast(\gamma_1)(t)=f\circ \gamma_1(t)=(2t,0)=2\gamma_1(t);\,\, f_\ast(\gamma_2)(t)=f\circ \gamma_2(t)=(0,-3t)=-3\gamma_2(t).$$

Thus $f_\ast$ takes the horizontal coordinate loop $\gamma_1$ and streches it by a factor of two so that $f_\ast(\gamma_1)$ is still horizontal but wraps around twice. Similarly $f_\ast$ takes the vertical coordinate loop $\gamma_2$, stretches it by a factor of three so that $f_\ast(\gamma_2)$ is still vertical, and also reverses its orientation. In terms of the isomorphism $H_1(\mathbb{T}^2;\mathbb{Z})\cong \mathbb{Z}^2$, $f_\ast$ is exactly the matrix $\begin{pmatrix}2&0\\0&-3\end{pmatrix}$ by which the original toral endomorphism $f$ is given.

(It might be a good idea to relate the four loops in Example 1 to each other by way of toral endomorphisms; e.g. what is an example of a toral endomorphism $g$ with $g_\ast(\gamma_1)=-\gamma_1+\gamma_2$ etc.)

Finally note that if $f$ and $g$ are both continuous self-maps of $\mathbb{T}^q$, then so is $f\circ g$, and $(f\circ g)_\ast(\gamma)=(f\circ g)\circ \gamma=f\circ (g\circ \gamma)=f_\ast(g_\ast(\gamma))=(f_\ast\circ g_\ast)(\gamma)$, so that the process of "inducing on homology" preserves compositions of functions. Thus any group action $\phi: \mathbb{Z}^p\to\operatorname{Homeo}(\mathbb{T}^q)$ by homeomorphisms (and in particular any group action $\phi: \mathbb{Z}^p\to\operatorname{Aut}_{\text{Lie}}(\mathbb{T}^q)$ by Lie group automorphisms) induces a group action $\phi_\ast: \mathbb{Z}^p \to \operatorname{Aut}_{\text{Ab}}(H_1(\mathbb{T}^q;\mathbb{Z}))$ by group automorphisms.

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  • $\begingroup$ Thank you so much. This is a very clear answer. $\endgroup$ Dec 28, 2021 at 17:20
  • $\begingroup$ @JoséLuisCamarilloNava I'm glad it was useful. Like I asked above I would be interested in having a look at the lecture notes you are reading, if they are publicly available. $\endgroup$
    – Alp Uzman
    Dec 28, 2021 at 17:32

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