# Showing that the set of semi-orthogonal matrices is a $C^\infty$ submanifold

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.

• In your title you say that $S_{nk}$ is a group. It can be such only for addition... do we agree ? – Jean Marie Apr 27 '16 at 21:23
• 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? – moran Apr 27 '16 at 21:38
• Neither it is stable by addition. My bad: I made a mistake. – Jean Marie Apr 27 '16 at 21:42

## 2 Answers

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 $nk-k(k+1)/2$.

• 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 $C^\infty$.) – moran Apr 27 '16 at 21:31
• 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 $d_{x}f$ – studiosus Apr 28 '16 at 7:53
• 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. – moran Apr 28 '16 at 10:15
• 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, $d_{X}Y=((X+tY)^{T}(X+tY)-I)'(0)=X^{T}Y+Y^{T}X$. – studiosus Apr 28 '16 at 10:33
• Ok, that makes sense. Thanks again. – moran Apr 29 '16 at 23:31

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.

• 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? – moran Apr 27 '16 at 21:35
• 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. – Jean Marie Apr 27 '16 at 21:39
• 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? – moran Apr 27 '16 at 21:43
• 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. – Jean Marie Apr 27 '16 at 22:05
• Edit added: is it OK now ? Thanks for having pointed this error to my attention. – Jean Marie Apr 27 '16 at 22:29