2
$\begingroup$

For $k, n \in \mathbb{N}$ with $k ≤ n$, we define

$$S_{n, k} = \{X \in \mathbb{R}^{n \times k}: X^t X = I_k\}$$

where $I_k$ is the identity matrix of rank $k$.

I want to prove that $S_{n, k}$ is a $C^\infty$-submanifold of the $\mathbb{R}^{n \times k}$, and want to find it's dimension. Using that, I want to conclude that $S_{n, k}$ is compact.

Now I thought about considering the mapping $f: \mathbb{R}^{n \times k} \to \mathbb{R}^{k \times k}, X \mapsto X^t X - I_k$. Then $S_{n, k}$ would be exactly the set of matrices that $f$ sends to $0$, and $\mathbb{R}^{k \times k} \cong \mathbb{R}^{k^2}$ has dimension $k^2$ whereas $\mathbb{R}^{n \times k}$ is of dimension $nk$, hence, the dimension of $S_{n, k}$ should be $(n - k)k$, I believe?

However, I still need to show that $f$ is not only continuously differentiable once and has a Jacobean matrix of proper rank (which would give me that $S_{n, k}$ is a $C^1$ manifold, I believe, since every $X \in S_{n, k}$ is a regular point then), but infinitely often, in order for $C^\infty$, do I? (Or is there an easier way?) And I don't really know how to get started with that.

Alternatively, I know I could also find a chart and show that it's $C^\infty$, but I don't know if that's easier either.

Thanks in advance.

EDIT: It also came to my mind that a chart might even be the more handier thing, since I'm supposed to conclude that $S_{n, k}$ is compact. Because the way the exercise is proposed makes it look like I'm supposed to use a topology argument, e.g. something like "$S_{n, k}$ is compact as the image of a compact space regarding a homeomorphism/diffeomorphism". I'm not sure though, and I'm of course open to all different kinds of approaches.

$\endgroup$
  • $\begingroup$ In your title you say that $S_{nk}$ is a group. It can be such only for addition... do we agree ? $\endgroup$ – Jean Marie Apr 27 '16 at 21:23
  • $\begingroup$ Corrected it to "set" instead of group, thanks. Right, it can't be a group regarding multiplication because the dimensions aren't right (except if $k = n$). But it can't be a group with addition either because the $0$-matrix (which would need to be the neutral element) isn't contained? $\endgroup$ – moran Apr 27 '16 at 21:38
  • $\begingroup$ Neither it is stable by addition. My bad: I made a mistake. $\endgroup$ – Jean Marie Apr 27 '16 at 21:42
1
$\begingroup$

Your approach is good, but it is important to choose the appropriate target set for $f$. We better consider $f$ as a map between $\mathbb{R}^{n\times k}$ and the set of symmetric $k\times k$ matrices (which has dimension $k(k+1)/2$). This will allow you to show that $0$ is a regular value of $f$, as follows:

You can check that the derivative $d_{X}f(Y)$ is given by $X^{T}Y+Y^{T}X$. We choose $X\in f^{-1}(0)$ and show that $d_{X}f$ is surjective (for this, the choice of the target set of $f$ is important). Take a symmetric $k\times k$ matrix $B$. Then we have
$d_{X}f(\frac{1}{2}XB)=\frac{1}{2}X^{T}XB+\frac{1}{2}B^{T}X^{T}X=\frac{1}{2}B+\frac{1}{2}B^{T}=B$.
Hence, $S_{n,k}$ will be a submanifold of $\mathbb{R}^{n\times k}$ of dimension $nk-k(k+1)/2$.

$\endgroup$
  • $\begingroup$ Thanks, that definitely helps. I wonder though, this only proves that $S_{n, k}$ is a $C^1$-manifold but not $C^\infty$ though, does it? Could we just give an inductive argument like "since the $n$-th derivative is a linear combination of products of the shape $X^t Y$ for some matrices $X, Y$, the $n+1$-th derivative is aswell" (just a bit more fleshed out) and thereby conclude that $f$ is infinitely often differentiable? (Which I believe I need for $S_{n, k}$ being in $C^\infty$.) $\endgroup$ – moran Apr 27 '16 at 21:31
  • $\begingroup$ For sure, $f$ is infinitely often differentiable: the derivative $d_{X}f$ is a linear map, hence when you differentiate it you again obtain $d_{X}f$ (see for instance this post: math.stackexchange.com/questions/392237/…). By induction, it then follows that the $k$-th derivative of $f$ at $X$ is also $d_{x}f$ $\endgroup$ – studiosus Apr 28 '16 at 7:53
  • $\begingroup$ Thanks again. There is only one thing that still gives me troubles: can you maybe provide a link or tip on how we can actually get the first derivative, i.e. how I can check that $d_Xf(Y)$ is given by $X^t Y + Y^T X$? I'm not that familiar with differentiating matrix mappings. $\endgroup$ – moran Apr 28 '16 at 10:15
  • 1
    $\begingroup$ If you want to find $d_{X}f(Y)$, you take any curve in $\mathbb{R}^{n\times k}$ that goes through $X$ at time zero with tangent vector $Y$. For instance, take $\alpha:t\mapsto X+tY$. Then $d_{X}Y$ is the velocity vector of the image curve $f\circ\alpha$ at time $0$. Hence, $d_{X}Y=((X+tY)^{T}(X+tY)-I)'(0)=X^{T}Y+Y^{T}X$. $\endgroup$ – studiosus Apr 28 '16 at 10:33
  • $\begingroup$ Ok, that makes sense. Thanks again. $\endgroup$ – moran Apr 29 '16 at 23:31
1
$\begingroup$

You don't need this result for proving that your set is compact.

In fact, set $S_{n,k}$ is defined by a system of $k^2$ equations

$$(E_{pq}:) \ \ \ \sum_{i=1}^n x_{pi}x_{qi}=\delta_{pq}$$

(where $\delta_{pq}$ is the Kronecker symbol: $1$ if $p=q$, $0$ otherwise).

Edit (following a remark of Moran pointing to a deficiency in my reasoning): In particular, if we add the constraints corresponding to the cases $\delta_{pp}=1$, we obtain the fact that the sum of the squares of all entries of $X$ is $1+1+...1=k$, proving that the set $S_{n,k}$ is included into the sphere $S$ of $\mathbb{R}^{n \times k}$ with radius $\sqrt{k}$, which is a compact set (it is a characteristic of finite dimensional spaces). Remark: using Frobenius norm, this constraint could be written $\|X\|_F^2=k$.

Each constraint $(E_{pq})$ defines a subset $E_{pq}$ of $\mathbb{R}^{n \times k}$ expressed as a certain $f_{pq}^{-1}\{a\}$ ($a=0$ or $1$) where $f_{pq}$ is continuous (even much more than that...).

Thus, $E_{pq}$, being the reciprocal image of a compact set (here $\{0\}$ or $\{1\}$) by a continuous function from the compact set $S$ (defined upwards) to $\mathbb{R}$, is itself a compact set. Inverse image of a compact set is compact

The intersection of these compact sets $E_{pq}$ is a compact set: we have reached the final objective.

$\endgroup$
  • $\begingroup$ Thanks, that's a great approach indeed. I'm having a little trouble though to follow: why is it that the $E_{pq}$'s are compact? $f$ is defined on $\mathbb{R}^{n \times k}$, which I think is not a compact set? Wouldn't we need this for the argument in the linked thread, or am I missing something? $\endgroup$ – moran Apr 27 '16 at 21:35
  • $\begingroup$ No, you don't need it. For example $\mathbb{R}$ is not compact, but if we define the continuous function $f: \mathbb{R} \rightarrow \mathbb{R}$ by $f(x)=x^2$, the inverse set of the compact set [0,1] is $f^{-1}([0,1])=[-1,1]$ compact. $\endgroup$ – Jean Marie Apr 27 '16 at 21:39
  • $\begingroup$ I'm still a little confused. If we take something like $f: \mathbb{R} \to \mathbb{R}, x \mapsto 0$, then the preimage of $\{0\}$ is $\mathbb{R}$ and thereby not compact. Where is the difference to our functions so that we can use this preimage argument? $\endgroup$ – moran Apr 27 '16 at 21:43
  • $\begingroup$ You are right, I completely overlooked this aspect: we have to check that the preimage is included in a compact set (i.e. a compact subset of $C_{pq} \subset \mathbb{R}^{n \times k}$)... I think that taking for $C_{pq}$ the hypersphere could be the answer. I try to make a convenient edit to my post. $\endgroup$ – Jean Marie Apr 27 '16 at 22:05
  • $\begingroup$ Edit added: is it OK now ? Thanks for having pointed this error to my attention. $\endgroup$ – Jean Marie Apr 27 '16 at 22:29

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.