Action of $S^1$ on a vector space

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 ?

• The proper translation to English would be "one-parameter group of linear transformations", i.e. a continuous homomorphism $S^1 \to \operatorname{GL}(V)$. Commented Jan 13, 2022 at 16:48
• Thank you very much @Alex Provost for correcting me.
– Mira
Commented Jan 13, 2022 at 17:38

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.

• thank you so much sir for this wonderful explanation.
– Mira
Commented Jan 13, 2022 at 20:17
• @asma, my pleasure. :) Commented Jan 13, 2022 at 20:17

My guess is that (1) is a bit of a mis-translation(?). It should probably say it is a parametrized family of linear transformations, since each element of $$S^1$$ corresponds to a particular sort of rotation matrix.

For the second question, no, there is not. You could take any line in $$V$$ and rotate about that line, and that gives an action of $$S^1$$ on $$V$$.

• And, yes, every continuous linear action of $S^1$ on $\mathbb R^n$ looks like the given formula, for some choice(s) of constants, after choosing suitable coordinates. Commented Jan 13, 2022 at 17:14
• @paul garrett, how can we prove it (that every continuos linear action of $S^1$ on $\mathbb{R}^n$ is represented by this matrix ) ?
– Mira
Commented Jan 13, 2022 at 17:40
• @Alfred Yeager, thank you!! But I still how we get that matrix ? Let's fix a choice of linear action of $S^1$ on $V$, and let $\theta \in S^1$ then we get an isomorphism $T_\theta : V \rightarrow V$. Let's choose a basis, how can we obtain a matrix of this form ?
– Mira
Commented Jan 13, 2022 at 17:43
• Calculate in the basis of eigenvectors of $T_\theta$. There are two possibilities for the eigenvectors: they can have eigenvalue $1$, coming from the dimensions where $T_\theta$ does nothing, or they can have complex eigenvalues, corresponding to planes where $T_\theta$ acts by rotation. Commented Jan 13, 2022 at 17:55
• May be you need to explain why the action is a rotation to start with. A priori it's just $\mathbb S^1 \to GL(V)$. Commented Jan 13, 2022 at 18:40