3
$\begingroup$

First of all for $\mathbb{R}$ in my book it is written that:

"$GL(n,\mathbb{R})$ is an open subset of euclidian $n^2$-space and that is the topology is given. Matrix multiplication is given by polynomials in the coefficients by Cramer's rule and so is continuous. Thus, $GL(n,\mathbb{R})$ is a topological group"

Question: Do we actually need to know that $GL(n,\mathbb{R})$ is an open subset in $M(n,\mathbb{R})$? As I understand if we show that $M(n,\mathbb{R})\times M(n,\mathbb{R}) \longrightarrow M(n,\mathbb{R})$ is continuous and, similarly, inverse for $M(n,\mathbb{R})$ is continuous this will imply that restriction to the subset topology, i.e $GL(n,\mathbb{R})$, is continuous.

Then, as exercise it is given to show that the same holds for $GL(n,\mathbb{C})$ and $GL(n,\mathbb{H})$. Since we know that there exist injective homomorphisms from $\mathbb{H}$ to the matrix rings $M(2,\mathbb{C})$ and $M(4,\mathbb{R})$ respectively. We can use the sketch of the proof above to get that $GL(n,\mathbb{C})$ and $GL(n,\mathbb{H})$ are also topological groups since their injective homeomorphisms in $M(2n,\mathbb{R})$ and $M(4n,\mathbb{R})$ are topological groups. Proving that they are open will be same as proving that the image of $GL(n,\mathbb{H})$, for example, under injective homeomorphism is in $GL(4n,\mathbb{R})$ which is also just a question whether linear independence of rows in some martix over $\mathbb{H}$ implies independence of rows in matrix in $\mathbb{R}$ under injective homeomorphism. Then $GL(n,\mathbb{H})$ is open in $M(n,\mathbb{H})$ as inverse of intersection of $GL(4n,\mathbb{R})$ and image of $M(n,\mathbb{H})$ because $GL(4n,\mathbb{R})$ was already proven to be open and because we use subset topology.

Question: Is this path to prove seems right? I think I finished all steps in between, so I just want to check if it is correct. Also, same question: Do we really have to show that $GL(n,\mathbb{C})$ and $GL(n,\mathbb{H})$ are open subsets to conclude that they are topological groups?

Please add name of the articles if you know where I can find proof of these facts.

Thank you!

$\endgroup$

2 Answers 2

8
$\begingroup$

Here's how you can see that $GL(n, \mathbb R)$ is an open subset of $M(n, \mathbb R)$: Consider the map $M(n, \mathbb R) \to \mathbb R$ given by $A \mapsto \det A$. From the definition of the determinant, this map is given by a polynomial in the coefficients of $A$, so it is continuous. Moreover $GL(n, \mathbb R)$ is the preimage of $\mathbb R \setminus \{0\}$. These two facts together tell you that $GL(n, \mathbb R)$ is an open subset of $M(n, \mathbb R)$. It is not actually necessary to know this to show that $GL(n \mathbb R)$ is a topological group, but it is a useful fact nonetheless (for instance, it gives you a canonical manifold structure for $GL(n, \mathbb R)$).

Secondly, it is true that if you show that the matrix multiplication map $M(n, \mathbb R) \times M(n, \mathbb R) \to M(n, \mathbb R)$ is continuous, then its restriction to $GL(n, \mathbb R)$ will be continuous. However, this doesn't make your life easier. Instead, just think about what matrix multiplication is, and you'll see that the entries of $AB$ are polynomials in the entries of $A$ and $B$, and therefore matrix multiplication is a continuous map.

There is no inverse map for $M(n, \mathbb R)$ since not every matrix is invertible, but that's OK. Just use Cramer's rule to write out what $A^{-1}$ is, expressed in terms of the coefficients of $A$. It will follow immediately that the inverse map in $GL(n, \mathbb R)$ is continuous.

The proofs of these statements for $\mathbb C$ and $\mathbb H$ are only superficially different. The details are really, intrinsically, the same. It really is a good exercise for you to confirm this on your own.

$\endgroup$
3
  • $\begingroup$ I would not say that proofs for $\mathbb C$ and $\mathbb H$ are "superficially different." It requires much more things to prove that this are open subsets since $\mathbb H$ doesn't have determinant function, so that you have to do (maybe there is another way I don't know) injective homeomorphisms to $M(4n,\mathbb{R})$ There are also some non-trivial facts: block multiplication of matrices, prove that it is actually induced homeomorphism, prove that the image of $GL(n,\mathbb{H})$ is in $GL(4n,\mathbb{R})$. $\endgroup$
    – Vadim
    Jul 12, 2015 at 3:53
  • $\begingroup$ But, my question was mostly just to confirm that open set is not used in the proof that all these are topological groups. Secondly, I wanted to confirm the sketch of the proofs for C and H. Thank you for the response! $\endgroup$
    – Vadim
    Jul 12, 2015 at 3:55
  • 1
    $\begingroup$ I would still say the proofs for $\mathbb C$ are so similar that they really are the same proofs. Just replace $\mathbb R$ by $\mathbb C$ everywhere and you're done! For $\mathbb H$, you're right, things are more complicated due to the lack of a determinant function. This is quickly fixed by defining a continuous injective group homomorphism $GL(n, \mathbb H) \to GL(4n, \mathbb R)$, as you seem to be aware. Making the target $GL(2n, \mathbb C)$ makes it much easier to see that such a map is a continuous homomorphism, by the way. $\endgroup$
    – Alex G.
    Jul 12, 2015 at 4:04
2
$\begingroup$

I think your arguments are in the right direction in terms of using the real matrices to induce the needed structures on the quaternionic or complex matrices. As to why $GL(n, \mathbb{R})$ is an open set, I'm not sure I understand your question, but, this much I should share just in case you don't already know: $$ \text{det}: \text{GL}(n, \mathbb{R}) \rightarrow (-\infty, 0) \cup (0, \infty)$$ and the determinant $\text{det}(A) = \sum_{i_1 \cdots i_n} \epsilon_{i_1\cdots i_n}A_{1i_1} \cdots A_{ni_n}$ is a polynomial in the components of $A$ hence it is clear that $\text{det}$ gives a continuous function with respect to the usual metric topology on $M(n, \mathbb{R})$. But, note as I indicate above, the set of invertible matrices is the inverse image of the open set $(-\infty, 0) \cup (0, \infty)$ hence $\text{GL}(n, \mathbb{R})$ is open. Many other things flow from this simple observation as I think you know.

Of course, if you don't like this argument, you could try to show each point in the set of invertible matrices is an interior point directly from the definition of norm for matrices.

$\endgroup$
2
  • $\begingroup$ My question was mostly just to confirm that open set is not used in the proof that all these are topological groups. Secondly, I wanted to confirm the sketch of the proofs for C and H. $\endgroup$
    – Vadim
    Jul 12, 2015 at 3:57
  • $\begingroup$ I see, well, the whole set is open in the subspace topology essentially by definition of the subspace topology. But, knowing it is also open with respect to the ambient space of all matrices may be useful for certain questions. But, fair enough, as to show it is a topological group it is not needed. (but, this comment depends on context, some texts take manifolds to be subsets of $\mathbb{R}^n$, others abstract sets... my default is to agree with you here though). If you don't have it already, I enjoyed Tapp's book worldcat.org/title/matrix-groups-for-undergraduates/oclc/… $\endgroup$ Jul 12, 2015 at 13:33

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .