Action of $S^1$ on a vector space

Question

In page 21 of the article cohomologie équivariante et théorème de Stokes, there is a small paragraph which says :

Let $V$ be real vector space on which acts the circle $S^1$ by a group with a parameter of linear transformations $g(\theta)$ (with $g(\theta)= g(\theta + 2 \pi))$. Then $V$ is of dimension $n= 2l+r$ and has basis $(e_1,e_2,…,e_{2l},e_{2l+1},…,e_{2l+r}$ on which the action of $S^1$ is given by the matrix
$$\begin{pmatrix}
\begin{pmatrix}
\cos(a_1{\theta}) & -\sin(a_1{\theta}) \\
\sin(a_1{\theta}) & \cos(a_1{\theta})
\end{pmatrix} \\
& \ddots \begin{pmatrix}
\cos(a_l{\theta}) & -\sin(a_l{\theta}) \\
\sin(a_l{\theta}) & \cos(a_l {\theta})
\end{pmatrix} \\ & & \begin{pmatrix}
1 \\
& \ddots & 1
\end{pmatrix}
\end{pmatrix}$$

What does it mean that $S^1$ on $V$ by a group of a parameter of linear transformations ?

Why does the action of $S^1$ on $V$ is represented by the above matrix, is there a unique action of $S^1$ on a vector space ?

Answer 1

In addition to @AlfredYerger's accurate point about mis-translation for the terminology:

I'd describe "a continuous linear action of the group $S^1$ on a finite-dimensional real vector space $V$" as a continuous group homomorphism $\varphi:S^1\to GL_{\mathbb R}(V)$. (For infinite-dimensional $V$ the best definition is subtler, but irrelevant to the finite-dimensional case.)

It seems optimal to set-up application of spectral theorems... so we'd want $V$ to have an inner product $\langle,\rangle$. Ok, since $V$ is finite-dimensional, we can pick a basis and declare it orthonormal. But, further, we want $S^1$ to act by orthogonal transformations (hence, "normal"), so we average over the group $S^1$: $$ \langle v,w\rangle_{\mathrm new} \;=\; \int_{S_1} \langle \varphi(k)v,\varphi(k)w\rangle\;dk $$ where $dk$ denotes a rotation-invariant measure on $S^1$ giving it total mass $1$.

Then, by design, all the operators $\varphi(k)$ for $k\in S^1$ are orthogonal with respect to the new inner product, so the spectral theorem for normal operators applies. Further, they are mutually commuting (since $S^1$ is abelian), ... and eventually we end up with that matrix form.

(It may be simplest to pass through the complex number version of this discussion, and then translate back to reals at the end, but this is a matter of taste.)

In any case, it's not a trivial matter to see that every circle action has that form. But/and the key point is the finite-dimensional spectral theorem for normal operators.