Does anyone know any references that describe automorphism group $\operatorname{Aut}(\mathbb R^n\times \mathbb T^m)$? I searched for a long time but couldn't find it.
1 Answer
(Below we work with continuous endomorphisms / automorphisms, i.e. this is all taking place in the category of locally compact Hausdorff abelian groups and continuous homomorphisms.)
It's easier to describe the endomorphism ring first, then take its group of units. In general (in any additive category!) the endomorphism ring of $A \oplus B$ is the ring of "matrices"
$$\begin{bmatrix} \operatorname{End}(A) & \operatorname{Hom}(B, A) \\ \operatorname{Hom}(A, B) & \operatorname{End}(B) \end{bmatrix}.$$
Here we get lucky: $\operatorname{Hom}(\mathbb{T}, \mathbb{R}) = 0$ so one of the off-diagonal terms disappears. Using the fact that $\operatorname{End}(\mathbb{T}) \cong \mathbb{Z}$ and $\operatorname{Hom}(\mathbb{R}, \mathbb{T}) \cong \mathbb{R}$, we get that $\operatorname{End}(\mathbb{R}^n \oplus \mathbb{T}^m)$ is
$$\begin{bmatrix} M_n(\mathbb{R}) & 0 \\ M_{m \times n}(\mathbb{R}) & M_m(\mathbb{Z}) \end{bmatrix}.$$
Because our "matrices" are now "lower triangular" the group of units is just given by taking the group of units along the diagonal; we get that the automorphism group is
$$\begin{bmatrix} \operatorname{GL}_n(\mathbb{R}) & 0 \\ M_{m \times n}(\mathbb{R}) & \operatorname{GL}_m(\mathbb{Z}) \end{bmatrix}.$$
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$\begingroup$ I can never remember when continuity comes for free—are these topological automorphism/endomorphism groups, or the full abstract automorphism/endomorphism groups? $\endgroup$– LSpiceCommented Jul 22 at 3:32
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$\begingroup$ Thank you. If there is no reference, i may prove it . $\endgroup$– freeCommented Jul 22 at 3:36
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3$\begingroup$ @LSpice: these are topological. The full abstract automorphisms / endomorphisms are much larger, using the existence of a Hamel basis for $\mathbb{R}$ over $\mathbb{Q}$, etc. $\endgroup$ Commented Jul 22 at 3:53
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$\begingroup$ @free what else remains to prove? $\endgroup$ Commented Jul 22 at 5:28
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$\begingroup$ a relevant discussion: math.stackexchange.com/questions/2525355/… $\endgroup$ Commented Jul 22 at 6:03