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When can one continuously prescribe a unit vector orthogonal to a given orthonormal system?

$\def\RR{\mathbb{R}}$ This problem was solved by Whitehead, G. W., Note on cross-sections in Stiefel manifolds, Comment. Math. Helv. 37, 239-240 (1963). ZBL0118.18702. Such sections exist only in ...
David E Speyer's user avatar
46 votes

When can one continuously prescribe a unit vector orthogonal to a given orthonormal system?

Unless I'm missing something, I think that the hairy ball theorem states precisely that you cannot do this when $n = 3$ and $k = 1$. I'm not sure what happens for other values of $n$ and $k$.
Will Brian's user avatar
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22 votes

When can one continuously prescribe a unit vector orthogonal to a given orthonormal system?

The space of orthonormal $k$-frames in $\mathbb{R}^n$ is the Stiefel manifold $V(k, n) = SO(n)/SO(n - k)$. There is a natural $SO(k)$ action on $V(k, n)$ and the quotient is the oriented grassmannian $...
Michael Albanese's user avatar
21 votes
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Is there a lift of the q-Vandermonde identity to some geometric (motivic) identity for Grassmannians over $F_q$?

Assume $V$ is a vector space of dimension $m+n$, $M \subset V$ is a subspace of dimension $m$, and $N = V/M$. Let $p:V \to N$ be the projection. Consider the Grassmannian $X = Gr(k,V)$ and its ...
Sasha's user avatar
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17 votes

When can one continuously prescribe a unit vector orthogonal to a given orthonormal system?

Denoting the Stiefel manifold of orthonormal $k$-frames in $\mathbb{R}^n$ by $V(k,n)$ as in Michael Albanese's answer, what you are asking for is a section of the sphere bundle $$ S^{n-k-1}\to V(k+1,n)...
Mark Grant's user avatar
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17 votes
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Homology of the free loop space of a Grassmanian

The complex Grassmannian $Gr(2,4)$ can be realized up to homotopy as the homotopy fiber of the map $BU(2) \times BU(2) \rightarrow BU(4)$ which corresponds to the Whitney sum of two complex rank 2 ...
Aleksandar Milivojević's user avatar
16 votes
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What is the $\mathbb{Z}_2$ cohomology of an oriented grassmannian?

I was surprised to learn that the ring structure of $H^*({\rm Gr}^+(k,n);\mathbb{Z}_2)$ seems to be unknown, in general. The ring structure in the case $k=2$ is given in Korbaš, Július; Rusin, Tomáš, ...
Mark Grant's user avatar
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15 votes

Map of Grassmannians associated with a Veronese embedding

I don't know a reference, but here is a simple argument. Note that $G(2,V)$ (let me use linear notation) is a homogeneous space for $GL(V)$: $$ G(2,V) = GL(V)/P_2, $$ where $P_2$ is a parabolic. If $...
Sasha's user avatar
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14 votes
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Relation between the homotopy classes of maps on a torus, and maps on a sphere

One case in which you can establish a simple relationship is when $Y$ is a loop space. Suppose that $Y\simeq \Omega Z=\mbox{map}_*(S^1, Z)$. Then there is a bijection $[{\mathbb T}^d, Y]_*\cong [\...
Gregory Arone's user avatar
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Can one divide algebraic manifolds ? Make sense: $Gr(2,n)/ Gr(2,n+m) = P^{n-1}/P^{n+m-1} P^{n-2}/P^{n+m-2}$

Let me begin with a sketch of an answer to your question two, namely giving an interpretation of the cross-multiplied equality $$\text{Gr}(2,n)\mathbb{P}^{n+m-1}\mathbb{P}^{n+m-2}=\text{Gr}(2,n+m)\...
Daniel Litt's user avatar
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14 votes
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Grassmannians on a vector space without metric

$\DeclareMathOperator\Gr{Gr}$I discussed this in my thesis. Lemma: Every tangent space of the Grassmannian is a tensor product $T_P \Gr(k)=P^* \otimes (E/P)$; these isomorphisms are invariant under ...
Ben McKay's user avatar
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13 votes
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Planes in Lagrangian Grassmannians

This is, indeed, true. To prove this, assume we have an embedding $\mathbb{P}^2 \to \operatorname{LGr}(V)$ (where $V$ is a symplectic vector space). Let $U \subset V \otimes \mathcal{O}$ be the ...
Sasha's user avatar
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12 votes

What is the amplituhedron?

There is now an AMS Notices article whose title is the same as the title of this question (and which therefore may be enlightening to anyone on this page): http://www.ams.org/journals/notices/201802/...
Sam Hopkins's user avatar
12 votes

Vector bundles on Stein manifolds

As has been established in the comments, the answer to your question is yes. It is a special case of a general result known as the Oka principle which has been strengthened over time. The key is that $...
Michael Albanese's user avatar
12 votes

Why are Lagrangian subspaces in a symplectic vector space interesting?

In symplectic linear algebra this is perhaps not completely clear. However, if one passes to symplectic geometry then the Lagrangean submanifolds indeed play a dominant role. This is in some sense ...
Stefan Waldmann's user avatar
11 votes

rational cohomology of finite real grassmannian

I could not find the explicit formulas in the Algebraic models book (they seem to only do infinite Grassmannians and Stiefel varieties) or Mimura-Toda (they do the complex and symplectic case but not ...
Matthias Wendt's user avatar
11 votes
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Quotients of Grassmannians

I am writing this as an answer because the comments are already too long. In the following I am incredibly pedantic, because there seems to be endless possibility for confusion with the several ...
11 votes
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Fundamental domain for two Grassmannians

Your argument about the dimension of the quotient doesn't take into account that there may be elements of the orthogonal group that don't do anything to the pair $(\pi_1,\pi_2)$. For example, if $2k&...
Robert Bryant's user avatar
10 votes

Canonical bundle of the Lagrangian Grassmannian

Here is a group theoretical solution which makes it is easy to compute the canonical bundle of any generalized flag variety. The Lagrangian Grassmannian is the homogeneous space $G/P$ where $G=Sp(2n)$...
Friedrich Knop's user avatar
10 votes

How few $k$-dimensional subspaces of $V$ are enough to have a complement to each $n-k$-dimensional subspace?

If $F$ is algebraically closed then $d(n,k)=k(n-k)+1$. For $W\subset V$ of dimension $k$, write $X_W\subset Gr(V,n-k)$ for the set of $n-k$-dimensional subspaces of $V$ that intersect $W$ non-...
Julian Rosen's user avatar
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10 votes
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Schubert calculus expressed in terms of the cotangent space of the Grassmannians

The tangent space to the Grassmanian corresponds to the following representation of $U(r)\times U(n-r)$, call it $\rho$: it is the $r\times (n-r)$ matrices, with $U(r)$ acting on the left and $U(n-r)$ ...
Anton Mellit's user avatar
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10 votes
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Is there a geometric interpretation of skew Schur functions?

This is discussed in Stanley's paper Some combinatorial aspects of the Schubert calculus. Corollary 3.7 says that under the natural isomorphism given by the Borel presentation of $H^*(G/P)$ which ...
Michael Joyce's user avatar
10 votes
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Is it possible that the GHKK canonical basis for cluster algebras is the Lusztig/Kashiwara dual canonical basis?

I think there is good reason to think the answer is "no". In rank 2, the theta basis agrees with the greedy basis (arXiv:1508.01404). Greedy basis elements are indecomposable positive elements (see ...
Hugh Thomas's user avatar
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10 votes
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The dimension of the Grassmannian cohomology ring $H^*(\mathrm{Gr}_{n,d})$ and the fundamental $\frak{sl}_n$-representation $V_{\pi_d}$

This is very standard. For a compact complex variety admitting a cell decomposition, the (co)homology is the free Abelian group generated by the cells (over $\mathbb{C}$ there is no room for the ...
Vladimir Dotsenko's user avatar
9 votes

Is it possible that the GHKK canonical basis for cluster algebras is the Lusztig/Kashiwara dual canonical basis?

Quoting from Geiss, Christof; Leclerc, Bernard; Schröer, Jan, Preprojective algebras and cluster algebras., ...
Jan Grabowski's user avatar
9 votes
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Is the representation of $GL_n(\mathcal{O})$ in functions on Grassmannian multiplicity free?

Yes, this is due to Hill: Hill, Gregory, On the nilpotent representations of (GL_ n({\mathcal O})), Manuscr. Math. 82, No. 3-4, 293-311 (1994). See especially Corollary 3.2. This was generalised and ...
A Stasinski's user avatar
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8 votes

How few $k$-dimensional subspaces of $V$ are enough to have a complement to each $n-k$-dimensional subspace?

Here is a particular case showing that $d(n,k)=d_K(n,k)$ definitely depends on the field $K$ in general, beyond pathologies of finite fields: $d_{\mathbf{R}}(4,2)=4$ (although $d(4,2)=5$ in the ...
YCor's user avatar
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8 votes

Is the Lie quadric $Q^3$ isomorphic to the Lagrangian Grassmannian $\operatorname{LG}(2,4)$?

Replying to this somewhat late but I think there is important context to this. The Lagrangian Grassmannian and the quadric are both examples of R-spaces (symmetric R-spaces in particular) as such they ...
Callum's user avatar
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8 votes
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Jumping conics in Grassmannians

These conics are exactly conics contained in some $$ \mathbb{P}^{n-1} \subset Gr(1,n), $$ that parameterizes all lines in $\mathbb{P}^n$ passing through a fixed point.
Sasha's user avatar
  • 37.6k
8 votes

The dimension of the Grassmannian cohomology ring $H^*(\mathrm{Gr}_{n,d})$ and the fundamental $\frak{sl}_n$-representation $V_{\pi_d}$

Harry Tamvakis gives an explanation of the relation between Grassmannian cohomology and representations of SL_n in The connection between representation theory and Schubert calculus. One fundamental ...
Oliver's user avatar
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