# On when quotients by a group action is a fibration or not, grassmanians

Let $$V$$ be a vector space over $$\mathbb R$$ with $$\text{dim }V\geq k$$ and let $$M = \text{Mon}(\mathbb R^k, V)$$ denote the space of monomorphisms (injective linear maps) of $$\mathbb R^k$$ into $$V$$.

Then there is a quotient map $$M \rightarrow Gr(k,V)$$. Is this a fibration? I think the map $$M \rightarrow Gr(k,V)$$ is equivalent to the quotient of $$\text{Mon}(\mathbb R^k, V)$$ by $$GL(k)$$ acting on the right by precomposition? I can't find any info on when quotients of smooth manifolds by a smooth action of a Lie group is a fiber bundle/fibration. Most theorems require that the smooth manifold be compact (which isn't true for my case).

Let's just assume that the map is a fibration for now... what is its fiber? I'm pretty sure that it is $$GL(k)$$ since the action of $$GL(k)$$ on $$M$$ is free. If $$A,B \in GL(k)$$ and $$\phi:\mathbb R^k \rightarrow V$$ is a monomorphism then obviously $$\phi A = \phi B$$ implies that $$A = B$$ since $$\phi$$ is mono.

I will accept any answer that proves whether $$M \rightarrow Gr(k,V)$$ is a fibration/fiber bundle or not but I am also curious about what the general theory of actions of lie groups on manifolds has to say and if there are general theorems that my question is a special case of!

Thanks in advance for any help!

According to Moishe Kohan's answer here, if you have a free and proper action of a Lie group $$G$$ on a manifold $$M$$ (no assumptions about compactness on either $$G$$ or $$M$$), then the quotient map $$M\rightarrow M/G$$ is a $$G$$-principal bundle.

As you noted, there is a free action of $$Gl(k)$$ on $$M$$ given by precomposing. So let's prove that this action is proper, meaning that the map $$\phi:G\times M\rightarrow M\times M$$ given by $$\phi(g,m)\rightarrow (gm, m)$$ is proper.

To that end, first note the following, which confirms your guess that the fiber above any point is a copy of $$Gl(k)$$:

Proposition Given two maps $$f_1,f_2\in M$$, there is a $$g\in GL(k)$$ with $$f_1 \circ g = f_2$$ iff the image of $$f_1$$ is the same as the image of $$f_2$$. Further, if there is such a $$g$$, it is unique.

Proof: First, since $$g:\mathbb{R}^k\rightarrow \mathbb{R}^k$$ is an isomorphism, it's surjective so $$f_2 = f_1\circ g$$ has the same image as $$f_1$$.

Conversely, if $$f_2$$ and $$f_1$$ have the same image, define $$g$$ as follows. For each standard basis vector $$e_i$$ of $$\mathbb{R}^k$$, there is a unique element $$v_i\in \mathbb{R}^k$$ with $$f_1(v_1) = f_2(e_i)$$. Let $$g$$ be the matrix whose $$i$$th column is $$v_i$$. Then $$f_1(g(e_i)) = f_1(v_i) = f_2(e_i)$$, so $$f_1\circ g = f_2$$.

Finally, if both $$g, g'$$ satisfy $$f_2 = f_1 \circ g = f_1 \circ g'$$, then $$f_1 \circ g' g^{-1} = f_1$$. Since you've already noted the action is free, this foces $$g'g^{-1} = I$$, that is, $$g' = g$$.$$\square$$

We will also need the following lemma.

Lemma: Suppose $$f_i$$ is a sequence in $$M$$ and $$f_i\rightarrow f$$. Suppose $$v_i$$ is a sequence of vectors in $$\mathbb{R}^k$$ with $$f_i v_i\rightarrow 0$$. Then $$v_i\rightarrow 0$$.

Proof: Pick a background inner product on $$V$$ and use the usual inner product on $$\mathbb{R}^k$$. Write each $$v_i\in \mathbb{R}^k$$ in polar form: $$v_i = r_i x_i$$ with $$x_i$$ on the unit sphere $$S^{k-1}$$ in $$\mathbb{R}^k$$ and $$r_i = |v_i|$$. Let us assume that $$v_i$$ does not converge to $$0$$, which means there is an $$\epsilon \geq 0$$ with the property that $$r_i \geq \epsilon$$ for some infinite set of $$i$$s. Abusing notation, restrict to the subsequence of these $$i$$s and call the new sequence $$v_i$$.

Because $$r_i f_i(x_i) = f_i(v_i)\rightarrow 0$$ and $$r_i$$ does is bounded below, we must have $$f_i(x_i)\rightarrow 0$$. The $$x_i$$ all live on sphere, which is compact, so some subsequence (again called $$x_i$$) must converge to $$x\in S{k-1}$$. Then $$f_i(x) = f_i(x-x_i) + f_i(x_i)$$. Now, $$\lim_{i\rightarrow\infty} f_i(x_i) = 0$$ by hypothesis. Further, $$\lim_{i\rightarrow \infty} f_i(x-x_i) = 0$$. To see this, note that there is a universal bound $$K$$ for $$\{f_i(v)|v\in S^{k-1}\}$$ owing to the fact that $$f_i\rightarrow f$$. Then $$|f_i(x-x_i)|\leq K|x-x_i|\rightarrow 0$$ as $$i\rightarrow \infty$$.

In short, $$\lim_{i\rightarrow\infty} f_i(x) = 0$$. It now follows that $$f(x) = 0$$, which is a contradiciton because $$f$$ is injective and $$x\in S^{k-1}$$ so $$x\neq 0$$. $$\square$$

Now, let's show the action is proper. So, let $$f_i$$ and $$g_i$$ be sequences in $$M$$ and $$G$$ and assume that both $$f_i\rightarrow f\in M$$ and $$h_i:=g_i\cdot f_i\rightarrow h\in N$$. We must show that some subsequence of the $$g_i$$ converges in $$G$$.

By the above proposition, the image of $$f_i$$ is the same as the image of $$h_i$$. Again by the proposition, it follows that the image of $$f$$ and $$h$$ coincide, so there is a unique $$g\in G$$ with $$h = f\circ g$$.

We claim there is a subsequence of $$g_i$$ converging to $$g$$. To that end, let us suppose for a contradiction that there is a neighborhood of $$g$$ which contains none of the $$g_i$$. Thus, there is some $$\epsilon > 0$$ with the property that the matrix $$g - g_i$$ has at least one entry with is $$\geq \epsilon$$. As there are only a finite number ($$k^2$$) of entries of a matrix in $$Gl(k)$$, there is at least one entry (say, in row $$a$$, column $$b$$), for which a subsequence of the $$g_i$$s (for which I'll abuse notation and call the subsequence $$g_i$$) satisfy $$|(g-g_i)_{ab}|\geq \epsilon$$.

Now, we know that $$f_i \circ g_i \rightarrow f\circ g$$. So, $$(f\circ g - f_i \circ g_i)(v)\rightarrow 0$$ for any $$v\in \mathbb{R}^k$$. Now, $$f\circ g - f_i \circ g_i = f\circ g - f_i\circ g + f_i\circ g - f_i\circ g_i = (f-f_i)\circ g + f_i\circ (g- g_i)$$ and $$f_i\rightarrow f$$. This implies that $$(f-f_i)(g v)\rightarrow 0$$ for any $$v\in \mathbb{R}^k$$. This, now, implies that $$f_i\circ (g- g_i)(v)\rightarrow 0$$ for any $$v\in \mathbb{R}^k.$$

However, let $$v = e_a$$ and set $$w_i = (g-g_i)(e_a)\in \mathbb{R}^k$$. Then $$|w_i|\geq \epsilon$$ for every $$i$$, because the $$b$$-th entry of $$w_i$$ is, in absolute value, larger than or equal to $$\epsilon$$. On the other hand, $$f_i(w_i)\rightarrow 0$$. By the lemma, this implies $$w_i\rightarrow 0$$, which contradicts the fact that $$|w_i|\geq \epsilon$$.

Having reached a contradiction, we conclude that every open set about $$g$$ contains a $$g_i$$, so there is a subsequence of the original $$g_i$$ sequence converging to $$g$$.

• I'm using the characterization of properness given by Jack Lee here:math.stackexchange.com/questions/987038/… He says it can be found in his textbook, but I long ago loaned my copy to a student and never got it back so I'm not sure where. Mar 28 '20 at 18:10
• Thank you for such a great in-depth answer! Mar 31 '20 at 10:33