Why is the set of automorphism of a n-dimensional torus T (also denoted $T^n$) is $GL(n,\mathbb{Z})$ i.e the set of invertible matrices with integral coefficients?

In the book by Brocker and Dieck- Representations of Compact Lie Groups, the crucial step for proving this goes as follows-

If there is a homomorphism $\phi:T^n\to S^1$, then it induces the diagram

$$\require{AMScd} \begin{CD} \mathbb{Z}^n @>{\hookrightarrow}>> \mathbb{R}^n @>{D\phi}>> \mathbb{R}\\@VVV @VVV @VVV \\ \mathbb{Z} @>{\hookrightarrow}>> T^n @>{\phi}>> S^1 \end{CD}$$ which gives the result that $$D\phi(v_1,...,v_n)= \sum n_iv_i$$ where $n_i \in \mathbb{Z}$; where the vertical maps are projections.

Q.1. How is the above formula for $D\phi$ inferred?

Other than that argument, I have managed to show that $GL(n,\mathbb{Z})\subset Aut(T^n)$. But how to show the other inclusion? That is,

Q.2. Given an automorphism $\phi$ of $T^n= \mathbb{R}^n/\mathbb{Z}^n$, show that there exist a map $\Phi$ $\in$ $GL(n,\mathbb{Z})$ such that $\pi \circ\Phi = \phi$ , where $\pi:R^n\to T^n$ is the projection.

Thanks in advance!

  • $\begingroup$ Do you understand what the vertical arrows in this diagram represent? That's the first step... $\endgroup$
    – Lee Mosher
    Commented Apr 11, 2017 at 11:49
  • $\begingroup$ Lee Mosher : Yes. I am familiar with the concept of exponential map in a Lie group. $\endgroup$ Commented Apr 11, 2017 at 20:52
  • $\begingroup$ Well, those may be exponential maps of Lie groups, but it's also important to understand them as quotient maps in the category of Lie groups. $\endgroup$
    – Lee Mosher
    Commented Apr 11, 2017 at 22:14
  • $\begingroup$ Yes I'm familiar with them being covering maps, projection maps, etc. $\endgroup$ Commented Apr 16, 2017 at 6:55

3 Answers 3


Here we are talking about the automorphisms of a torus as a Lie group, that is we want group-automorphisms of $T^n$ which are smooth (actually continuous is enough as that will imply smoothness).

Let $T^n=\mathbb{R}^n/\mathbb{Z}^n$. Let's try to find all smooth homomorphisms $F$ from $T^m$ to $T^n$. Composing with the projection from $\mathbb{R}^m\to T^m$, $F$ induces a continuous homomorphism $\tilde F:\mathbb{R}^m\to T^n$. Every continuous map from $\mathbb{R}^m$ to a nice enough topological space lifts to its universal cover ($T^n$ is certainly nice enough) so $\tilde F$ lifts to a continuous map $\hat F$ from $\mathbb{R}^m$ to $\mathbb{R}^n$. We can ensure that $\hat F(0)=0$ and then $\hat F$ is a continuous homomorphism (this can be reduced to checking in the neighbourhood of zero). Then $\hat F$ is given by a real $m$-by-$n$ matrix. But $\hat F$ must take the lattice $\mathbb{Z}^m$ into $\mathbb{Z}^n$. Thus its entries are integers.

Conversely, an integer matrix induces a Lie group homomorphism from $T^m$ to $T^n$ given by matrix multiplication acting on vectors.

Composing these homomorphisms just multiplies their matrices. Therefore the endomorphisms of $T^n$ which are invertible come from precisely the invertible $n$-by-$n$ matrices.

  • $\begingroup$ "Then $\hat F$ is given by a real $m$-by-$n$ matrix. But $\hat F$ must take the lattice $\mathbb{Z}^m$ into $\mathbb{Z}^n$. Thus its entries are integers." How? Integral entries in a matrix will cause the matrix to fix the $\mathbb{Z}^n$, but how to show the converse? $\endgroup$ Commented Apr 16, 2017 at 13:37
  • $\begingroup$ Oh! I got it... $\endgroup$ Commented Apr 19, 2017 at 1:35
  • $\begingroup$ Why is $\hat{F}(0)=0$? $\endgroup$
    – Yuxi Han
    Commented Aug 12, 2019 at 23:20

To put it in a concise answer:

A.1: If for some $i$, $n_i \in \mathbb{R}$ but not in $\mathbb{Z}$, then the corresponding unit vector will map to $$D\phi(0,...,0,e_i,0,..0)=\sum_{i\neq j} n_j. 0 + n_i.e_i = n_i.e_i$$ which will not be sent to $\mathbb{Z}$. Hence the entries of the matrix must be integers.

A.2: (As detailed above) Any homomorphism $\phi \in Aut(T^n)$ can be lifted to $Aut(\mathbb{R}^n)$, because of the lifting property of universal cover. (Since $\mathbb{R}^n$ is the universal cover of $T^n$). Therefore $$Aut(T^n)\subset Aut(\mathbb{R}^n)$$ Hence the automorphism is given by a matrix.


Concerning your first sentence I want to add a further reference here, where we see that the automorphism group of $T^n=\mathbb{R}^n/\Lambda$, where $\Lambda$ is a lattice (here $\mathbb{Z}^n$) is isomorphic to the automorphisms of the lattice $\Lambda$, which is the group of linear maps $\mathbb{R}^n\rightarrow \mathbb{R}^n$ mapping $\Lambda$ to $\Lambda$. So the group is $$ Aut(T^n)=Aut(\mathbb{Z}^n)=GL_n(\mathbb{Z}). $$

  • $\begingroup$ Oh sorry, I meant why $Aut(T^n)=Aut(\mathbb{Z}^n)$, when they say "The automorphism group you describe is always equal to the group of automorphisms of the lattice " $\endgroup$ Commented Apr 11, 2017 at 21:48

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