Suppose $V, V'$ are subspaces of dimension $d$ of a vector space $X$. Then there is a subspace $W$ of $X$ of codimension $d$ such that $W \cap V = W \cap V' = { 0 }$. This can be proved by choosing an explicit basis for $X$ which contains a basis for $V$ and a basis for $V'$ and a basis for $V \cap V'$. On the other hand, there should be a nice way to do this without choosing a basis. Can anyone explain this?

Disclosure: This came up when I was doing a homework problem. However, I'm just going to use the non-basis-free approach when I write up my answer.

  • $\begingroup$ I'm not sure I'm following, so please let me know how that proof you mention goes. For instance, if X=R^4, V=gen{e1,e2,e3}, V'=gen{e2,e3,e4}, where ei's are the canonical base vectors, we wish to find a subspace W of dim 1 such that W \cup V = W\cup V' = \empty, how is this proved? $\endgroup$ – Weltschmerz Oct 3 '10 at 0:37
  • $\begingroup$ @Weltschmerz: take W = span{e_1+e_4}. In any case, Akhil, this probably follows from known facts about Grassmannians, but I don't immediately see an elementary way to do this in a basis-free manner. $\endgroup$ – Qiaochu Yuan Oct 3 '10 at 1:00
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    $\begingroup$ Let $c = \dim V \cap V'$. Then we can find a basis for $V \cap V'$ of $c$ elements. We can extend this in one direction to a basis for $V$, and in another direction for $V'$. The union is a basis for $V + V'$. We can complete this to a basis for $X$. So we can choose a basis for $X$ of the following form: $\{f_1, \dots, f_{d-c}, e_1, \dots, e_{d-c}, g_1, \dots, g_c, h_1, \dots, h_{n - 2d + c}\}$ where $n = \dim X$. The set consisting of the $f_i + g_i$ and the $g_j$ has cardinality $n-d$ and its span does not intersect $V$ or $V'$. $\endgroup$ – Akhil Mathew Oct 3 '10 at 1:01
  • $\begingroup$ @Qiaochu: Yes, this actually came out of my doing an exercise on computing the dimension of a Grassmannian. It's not important to have a basis-free proof for the exercise; it just would be interesting if there is one (which I was unable to find). $\endgroup$ – Akhil Mathew Oct 3 '10 at 1:03
  • $\begingroup$ @Akhil. Shouldn't it be "a set containing $f_i+e_i$ and $h_j$"? $\endgroup$ – Jyotirmoy Bhattacharya Oct 3 '10 at 1:55

Suppose first that $V+ V' = X$. Then the natural map $X/(V\cap V') \to X/V \times X/V'$ is an isomorphism. Choose a subspace of the target of this isomorphism that projects isomorphically onto each factor (i.e. the graph of an isomorphism between the two factors; such an isomorphism exists since the two factors have the same dimension). Its preimage under the natural map is a subspace of $X/(V\cap V')$ which meets each of $V$ and $V'$ trivially. Now choose any subspace $W$ of $X$ that projects isomorphically onto this preimage; this is then a subspace of $X$ that maps isomorphically onto each of $X/V$ and $X/V'$, and hence is a codimension $d$ subspace with the desired property.

In general (i.e. if $V + V' \neq X$) then the above gives a codimension $d$ subspace $W'$ of $V + V'$ meeting $V$ and $V'$ trivially. Choose $W''$ to be any subspace of $X$ which maps isomorphically onto $X/(V + V')$. The sum $W' + W''$ is then a codimension $d$ subspace of $X$ meeting $V$ and $V'$ trivially.

  • $\begingroup$ That does it. Thank you! $\endgroup$ – Akhil Mathew Oct 3 '10 at 13:54

So let $G$ be the space of all co-dimension d subspaces. ($G$ is naturally a manifold). I claim that if $W$ is any dimension $d$ subspace, the set of $X$ s.t. $X \cap W = 0$ is an open, dense subset of $G$. Open is easy. If you have $X_i$ each with non-trivial intersection with W and $X_i \to X$ then you can pick lines $L_i$ in $W$ intersect $X_i$ and find some limit (perhaps of a subsequence) so that $L_i \to L$. Then $L$ is in $W \cap X$.

To show dense is also not hard. You need to show that if $X$ intersects $W$, then you can change $X$ by a little so that it no longer does. This really isn't hard but it is annoying to do without ever picking a basis of anything. How about this. You can think of $X$ as the image of a map $Y \to Z$ ($Y$ is a space of dim $n-d$, $Z$ is your big space). $X$ intersects $W$ trivially iff the map $Y \to Z/W$ is an injection. Pick some $U$ so that $Z = U + V$. Then we can think of our map as $Y \to U+V$, and we want the map $Y \to U$ to be injective. But $Y$ and $U$ have the same dimension and it is easy to modify a map between such spaces by epsilon to make it a bijection (for example add a small multiple of some particular bijection).


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