# Homeomorphic to a vector space but not itself a vector space

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?

• The article about transport of structure might be of interest to you: en.wikipedia.org/wiki/Transport_of_structure – Vercassivelaunos Apr 21 at 7:49
• 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. – Alejandro Epelde Apr 21 at 7:50
• @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 at 7:53
• 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. – Alejandro Epelde Apr 21 at 7:57
• @AlejandroEpelde Excellent, that makes sense. Thank you – quixot Apr 21 at 8:00

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.
• @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 at 8:42
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.