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In Nash & Sen p.162, they show that the space of all positive definite symmetric matrices $C$, while not a vector space itself, is homeomorphic to the space of all symmetric matrices $S$ (which is a vector space), via the map

$$ s \rightarrow e^s \in C, s \in S $$

My question: can I not make $C$ into a vector space by defining addition of two elements as first adding the elements in $S$ and then exponentiating, i.e. $ e^a + e^b \equiv e^{a + b}$? Note this is not the usual multiplication of matrices. It seems to me that we then have an inverse ($e^{-a}$) and an identity element ($e^0$), that we inherit commutativity etc. from $S$, and we can define scalar multiplication in the usual way.

More generally, why is something homeomorphic to a vector space not itself a vector space?

Can someone tell me where I'm going wrong?

quixot
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    The article about transport of structure might be of interest to you: https://en.wikipedia.org/wiki/Transport_of_structure – Vercassivelaunos Apr 21 '21 at 07:49
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    The issue is that the already existing operations on $C$ do not make it into a vector space. It is not even closed under taking inverses. However as you point out, you can 'borrow' the vector space structure of $S$ using the bijection $s\mapsto e^s$. This works in general, but again the vector space structure you obtain may have little to do with any already existing one. – A Epelde Apr 21 '21 at 07:50
  • @AlejandroEpelde Ah! They use the proof in the context of showing that C is contractible (because vector spaces are). Does that mean that, given any set, it is contractible if I can make it into a vector space (even if the addition rule or whatever is not the usual one)? – quixot Apr 21 '21 at 07:53
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    That seems reasonable, so long as the resulting vector space operations are continuous with respect to some topology on your set. Which in this case they are since $s\mapsto e^s$ is a homeomorphism. – A Epelde Apr 21 '21 at 07:57
  • @AlejandroEpelde Excellent, that makes sense. Thank you – quixot Apr 21 '21 at 08:00

2 Answers2

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Here is a simpler example: $(-1,1)$ and $\Bbb R$ are homeomorphic; just consider the map$$\begin{array}{rccc}f\colon&(-1,1)&\longrightarrow&\Bbb R\\&x&\mapsto&\tan\left(\frac\pi2x\right).\end{array}$$However, $(-1,1)$ isn't automatically a vector space. It becomes a vector space if you define the addition by $x+y=f^{-1}\bigl(f(x)+f(y)\bigr)$ and if you define the multiplication by a scalar by $\lambda x=f^{-1}\bigl(\lambda f(x)\bigr)$. This also works in that situation that you described indeed.

José Carlos Santos
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  • Okay great thank you! So as long as I can create rules which make a set into a vector space (equivalently, find a homeomorphism to a vector space) I can use the properties of vector spaces? In the Nash & Sen example, showing that a space is contractible because vector spaces are. – quixot Apr 21 '21 at 07:56
  • @quixot This is because this set is homeomorphic to a vector space. In general, for cardinality reason, one can find a bijection from some set to some vector space (for example, $S^1$ and $\mathbb{R}$ are in bijection), which cannot lead to a homeomorphism. – Didier Apr 21 '21 at 08:42
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The set of positive semi-definite matrices has a structure on its own: it is a cone (The space of positive semidefinite $n \times n$ matrices is a cone). In particular, it is convex. See also the deep article here with its geometrical point of view.

Jean Marie
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