1-Torus as finite dimensional $\mathbb{R}$-vector space is one dimensional, yet not isomorphic to $\mathbb{R}$

Question

I know that the 1-torus, given by its presentation as rotation matrixes:

$\mathbb{T}=\{R_{\theta}=\begin{pmatrix} cos(\theta) & -sin(\theta) \\ sin(\theta) & cos(\theta) \end{pmatrix}: \theta \in \mathbb{R}\}$, forms a vector space over $\mathbb{R}$, with the following addition and scalar multiplication.

$R_{\theta} \oplus R_{\alpha} = R_{\theta}R_{\alpha}=R_{\theta + \alpha}$, this works because of sine and cosine sum of angles formulae.

And scalar multiplication given by $r\odot R_{\theta} = R_{r\theta}$.

My doubt is the following, I'm quite sure given a finite dimensional vector space $V$ over a field $\mathbb{K}$, we can say that:

$$(1)\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }V\cong \bigoplus_{j=1}^{dim(V)}\mathbb{K}$$

Where the isomorphism is given by mapping the scalar multiplying each basis element to it's own coordinate on the direct sum.

The thing here, is that we have an epimorphism given by:

$$\psi:\mathbb{R}\to \mathbb{T}$$ $${\theta}\mapsto R_{\theta}$$

This epimorphism has clearly a nontrivial kernel, given the periodicity of sine and cosine functions, and the space defined as $\mathbb{T}$ is clearly one dimensional.

Also, no linear mapping between these spaces can be ever an isomorphism, given that $\mathbb{T}$ is compact, and every linear function is continuous on $\mathbb{R}$.

How come this isn't a contradiction, am i missing something and $\mathbb{T}$ isn't really a vector space?

What I'm sure has to be true is, given a vector space $V$, with an ordered basis $\mathcal{B}=\{b_{i}\}_{i=1}^{n}$, and the mapping:

$$\phi: \bigoplus_{j=1}^{n}\mathbb{K} \to V$$ $$(\lambda_{i})_{i=1}^{n} \mapsto \sum\limits_{i=1}^{n} \lambda_{i}b_{i}$$

Is an epimorphism, so because of the first isomorphism theorem for modules, we can conclude:

$$V\cong \bigoplus_{j=1}^{n} (\mathbb{K}/Ker(\phi_{i}))$$

Where $\phi_{i}:\mathbb{K}\to V$, given by $\phi_{i}(\lambda)=\lambda b_{i}$.

Is this what I should think about when talking about finite dimensional vector spaces, or is indeed (1) true, and I'm missing something fundamental about the structure of $\mathbb{T}$, making it NOT a vector space of finite dimension over $\mathbb{R}$?

Answer 1

In fact, the $1$-torus fails to be a vector space over $\Bbb R$. Following the list of axioms given here, the torus fails "compatibility of scalar multiplication with field multiplication". Note for instance that $$ \frac 14 \odot (4 \odot R_{\pi/2}) = R_0 \neq (\frac 14 \cdot 4) \odot R_{\pi/2}. $$

Answer 2

The putative scalar multiplication map, $$r \odot R_{\theta} \mapsto R_{r \theta} ,$$ is not even well-defined.

The periodicity of $\sin, \cos$ impliy that $$R_{\theta + 2\pi} = R_\theta .$$ But taking (for notational convenience) $\theta = 2 \beta$ and symbolically applying the rule for the scalar multiplication map (i.e., temporarily not worrying about well-definedness) gives that $$\frac{1}{2} \odot R_{2 \beta + 2\pi} = R_{\beta + \pi} = R_\beta R_\pi = - R_\beta,$$ which does not coincide with $$\frac{1}{2} \cdot R_{2 \beta} = R_{\beta} .$$

Putting this a little more abstractly (and formally): The map $\pi : \theta \mapsto R_{\theta}$ is a quotient map and identifies $\Bbb T$ with the space $\Bbb R / \sim$, where $x \sim y$ iff $\pi(x) \leftrightarrow \pi(y)$.

• The addition operation $+$ of the real vector space $\Bbb R$ descends via $\pi$ to an operation on $\Bbb T$, namely, $\oplus$. It follows that $\oplus$ satisfies the usual axioms of the vector space addition, and in particular $(\Bbb T, \oplus)$ is a group (isomorphic to $SO(2)$). (In fact, $\pi$ is a group homomorphism $(\Bbb R, +) \to (\Bbb T, \oplus)$.)
• On the other hand, the scalar multiplication operation $\cdot : \Bbb R \times \Bbb R \to \Bbb R$ does not descend to a map $\Bbb R \times \Bbb T \to \Bbb T$: As the above computation shows, $\pi(r \cdot \alpha)$ is not independent of the choice of representative $\alpha$ of $R_\theta$ in $\pi^{-1}(R_\theta) = \{\theta + 2 \pi k : k \in \Bbb Z\}$. But this descent was how the map $\odot$ was characterized, so it is not well-defined.