# The space of direct decompositions

The Grassmannian $$\mathrm{Gr}(k, n) = O(n) / O(k)\times O(n-k)$$ describes all $$k$$-dimensional subspaces of $$\mathbb R^n$$. The product space $$S=\mathrm{Gr}(k, n)\times \mathrm{Gr}(n-k, n)$$ represents pairs of subspaces $$(U, V)$$.

Some of these pairs form a direct sum decomposition $$U\oplus V=\mathbb R^n$$, forming a subspace $$S' := \{(U, V) \in S\mid U\oplus V=\mathbb R^n \}.$$

I am quite sure that $$S'$$ is a known mathematical object, but I can't find any references on it. I checked N. Steenrod's Fiber Bundles, S. Watanabe's Algebraic Geometry and Statistical Learning Theory, T. tom Dieck's Representation theory and Googled "Grassmannian of pairs", "flag Grassmanian", "summand Grassmannian", but I haven't found anything.

Could you recommend me some references on this object?

• Here is something you might try : we know how to characterize these pairs. Choosing a basis $u_1,\ldots, u_k$ for $U$ and a basis $v_1,\ldots, v_{n-k}$ for $V$, $\det(u_1,\ldots, u_k,v_1,\ldots, v_{n-k}) \ne 0$ if and only if $U+V$ is a direct sum (and hence spans all of $\Bbb{R}^n$). This function is clearly not independent of choice of basis, but its vanishing locus is. So, you might define $S'\subset S$ to be the zero locus of this function which we could informally write as $(U,V)\mapsto \det(U,V)$. I haven't tried the computation, but I bet if you write it in local coordinates... Commented Mar 31, 2023 at 15:44
• you will find that it is a submanifold of $S$. Commented Mar 31, 2023 at 15:44
• Indeed! :) This is also how I handle these computationally (enforcing the absolute value of the determinant to be larger than a chosen $\epsilon$ for numerical stability). It's also possible to generate them using $GL(n)$ by writing $\mathbb R^n=\mathbb R^k \oplus \mathbb R^{n-k}$ and acting with a matrix $A=GL(n)$ (again, in practice with an $\epsilon$ constraint). However, I believe this is an important enough object that it has been studied before and may have quite interesting theory... Commented Mar 31, 2023 at 15:54
• Yet another possibility is to describe them using the Schubert variety and use the canonical angles between flats, but I still believe it's not the end of the story. Commented Mar 31, 2023 at 15:58
• Isn't it just $GL(n)/(GL(k)\times GL(n-k))$? Commented Apr 2, 2023 at 4:57

As stated in the comment by @Deane you can easily describe the space of interest as the homogeneous space $$GL(n,\mathbb R)/(GL(k,\mathbb R)\times GL(n-k,\mathbb R))$$. The argument for this is that $$GL(n,\mathbb R)$$ acts transitively on the spaces pairs of subspaces $$V,W\subset\mathbb R^n$$ of the right dimensions such that $$\mathbb R^n=V\oplus W$$ and for the "standard pair" $$V=\mathbb R^k$$, $$W=\mathbb R^{n-k}$$, the stabilizer subgroup is the subgroup $$GL(k,\mathbb R)\times GL(n-k,\mathbb R)$$ of block diagonal matrices.
Howewer this does not correspond to the "orthogonal" interpretation of the Grassmannian as $$O(n)/(O(k)\times O(n-k))$$ that you start from. Indeed that space of decompositions $$\mathbb R^n=V\oplus W$$ is not a homogeneous space of $$O(n)$$ (since (roughly speaking)) $$O(n)$$ presereves the "angle" between $$V$$ and $$W$$. The corresponding homogeneous space of $$O(n)$$ is the space of orthogonal decompositions $$\mathbb R^n=V\oplus_{\perp} W$$, but this coincides with the Grassmannian itself, since any $$k$$-dimensional subspace determines the decomposition $$\mathbb R^n=V\oplus V^\perp$$ and this is the only orthogonal decomposition with first factor $$V$$.
So to get the Grassmannian nicely into the picture, you should view it as $$GL(n,\mathbb R)/P$$, where $$P$$ is the subgroup $$\left\{\begin{pmatrix} A & B\\ 0 & C\end{pmatrix}:A\in GL(k,\mathbb R), C\in GL(n-k,\mathbb R)\right\}$$. Then you can view $$H:=GL(k,\mathbb R)\times GL(n-k,\mathbb R)$$ naturally as subgroup of $$P$$ and correspondinly, you get a projection $$G/H\to G/P$$ (defined by $$gH\mapsto gP$$) which maps a decomposition to its first summand. Indeed, this is a fiber bundle with standard fiber $$P/H$$ (which naturally is an affine space).