# Question regarding classification of subspace configurations up to isomorphism and classifying orbits

Let $$n\geq 1$$ and $$V=(V_0, V_1, ... ,V_n)$$ be a $$(n+1)$$-tuple of vector spaces.

Let $$\mathcal{U_n}(V)=\{(f_1,...,f_n)|f_i \in Hom_K(V_i, V_0), f_i$$ is injective, $$1\leq i\leq n\}$$

Set $$G(V ) := GL(V_0) × GL(V_1) × · · · × GL(V_n)$$

And define the map

$$η : G(V ) × \mathcal{U_n}(V ) → \mathcal{U_n}(V )$$ $$((g_0, . . . , g_n),(f_1, . . . , f_n)) \mapsto (g^{−1}_0 f_1g_1, . . . ,g^{−1}_0 f_n g_n)$$

The orbit of $$f=(f_1,...,f_n)$$ is $$\mathcal{O}(f):= \{η(g, f) | g ∈ G(V )\}$$

To classify the orbits in $$\mathcal{U_n}(V )$$ one uses the following.

A $$n$$-subspace configuration is a $$(n+1)$$-tuple $$(V, V_1, ..., V_n)$$, where $$V$$ is a vector space and $$V_1, ..., V_n$$ are subspaces of $$V$$.

Two $$n$$-subspace configurations $$(V, V_1, ..., V_n)$$ and $$(W, W_1, ..., W_n)$$ are isomorph if there exists an isomorphism $$f: V \to W$$ with $$f(V_i)=W_i$$ for all $$1\leq i \leq n$$

For $$V=(V_0, V_1, ..., V_n)$$ the map defined by

$$(f_1, . . . , f_n) \mapsto (V_0,im(f_1), . . . ,im(f_n))$$

induces a bijection between the set of all $$G(V)$$-orbits in $$\mathcal{U_n}(V)$$ and the set of isomorphy classes of $$n$$-subspace configurations $$(W, W_1, . . . , W_n)$$ mit $$dim(W) = dim(V_0)$$ and $$dim(W_i) = dim(V_i)$$ for $$1 ≤ i ≤ n$$.

My questions:

1. Does that mean that two $$n$$-subspace configurations are isomorph if the dimensions of their entries match? I suspect it is not enough because it needs to be the same isomorphism which "matches" them all. What exactly is the condition for the demanded isomorphism to exist?

2. By what relation exactly am I classifying the orbits? What relation must they stand in for them to be in the same class? Is it even given? Or am I looking for a relation they can be classified by?

3. How exactly would I go on to classify the orbits in $$\mathcal{U_n}(V)$$ for example in the cases of (a) $$V=(K^n, K^a, K^b)$$ with $$a+b \leq n$$, (b) $$V=(K^2, K, K, K)$$ and (c) $$V=(K^3, K, K, K, K)$$?

I would like some hints as I am endlessly confused and do not know where to start.

I have tried looking the problem up in the internet and did not find any resources. I am welcoming all further reading recommendations on this problem.

My idea for 3.a is to consider the $$2$$-subspace configuration $$(K^n, im(f_1), im(f_2))=(K^n, U_1, U_2)$$ because the images are subspaces of $$K^n$$ with the dimensions $$a$$ and $$b$$. Let $$dim(U_1 \cap U_2)=d$$. The ordered basis of $$U_1 \cap U_2$$ can be expanded to an ordered basis of $$U_1, U_2$$ or $$K^n$$.

$$\mathcal{B}_{U_1 \cap U_2}=(b_1, ..., b_d)$$ $$\mathcal{B}_{U_1}=(b_1, ..., b_d, u_1, ..., u_{a-d})$$ $$\mathcal{B}_{U_2}=(b_1, ..., b_d, v_1, ..., v_{b-d})$$ $$\mathcal{B}_{K^n}=(b_1, ..., b_d,u_1, ..., u_{a-d}, v_1, ..., v_{b-d})$$

I strongly suspect that this means: There is an isomorphism between two configurations $$(K^n, im(f_1), im(f_2))$$ and $$(K^n, im(f_3), im(f_4)) \iff dim(im(f_1) \cap im(f_2))=dim(im(f_3) \cap im(f_4))$$.

Is that a guess in the right direction?

I also see that (b), (c) can be seen as an instance the previous problem if two images match. Is this correct?

Not sure if this helps anyone, but this is an instance of the problem that Gelfand und Ponomarev solved for $$n=4$$ in 1970. However I have no access to their paper and I am guessing it is too high for me as I am in my first year of linear algebra.

1. No. As you noticed on your example on $(K^n,K^a,K^b)$, other than the dimensions, also the intersections of the subspaces are important. For example, if you take $v\in \mathbb R^2$ and $V$ the span of $v$, you get that $(\mathbb R^2,V,V)$ is not equivalent to $(\mathbb R^2,V,V^\perp)$, since $V\ne V^\perp$. ($\perp$ indicates the orthogonal with respect to the standard scalar product of $\mathbb R^2$). The condition to for such an isomorphism to exist in fact, is that the dimensions of all the intersections of subspaces coincide. It's easy to be convinced that it is a necessary condition since an isomorphism can't change the dimensions. It's quite more difficult to prove it's sufficient, but the reasoning is similar to what you did to find the configurations for $(K^n,K^a,K^b)$.
2. As you said, the orbits correspond to the class of isomorphisms of $n$-spaces inside $V_0$. Literally, two elements $f,f'$ of $\mathcal{U_n}(V)$ are in the same orbit if and only if the two $n$-spaces $(V_0,im(f_1), . . . ,im(f_n))$ and $(V_0,im(f'_1), . . . ,im(f'_n))$ are isomorphic. this is the relation.
3. As said, the dimensions of all the intersections of subspaces must coincide. you did well for case $(a)$. In the other two cases you can see that the intersection of the subspaces may have only dimension 1 or 0, so they coincide or they have intersection $\{0\}$. In case $(b)$ you obtain 5 possible outcomes, while in case $(c)$ you obtain 15 outcomes (In general, if you have $(K^n,K,K,K,\dots,K)$ with $n>1$, and $k$ subspaces, the number of possible outcomes is the $k-th$ Bell Number)
Edit: what I said works only for infinite fields $K$. In the case of a finite fields, things become much more... interesting