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Jan
13
comment embedding of quaternionic projective spaces
The embedding can be geometrically described as follows. Let $V$ be the real vector space of $3\times3$ Hermitian quaternionic matrices with fixed (real) trace, say $1$. Note that $\dim V=14$. $\mathbb HP^2$ embeds into $V$ by mapping each quaternionic line in $\mathbb H^3$ to the matrix representing the corresponding orthogonal projection onto it. The image of the embedding in $V$ consists of the idempotent matrices, and it is also an orbit of the action of the group $Sp(3)$ on $V$ by conjugation. The image sits in the unit sphere of $V$, so it can be stereographically projected to $R^{13}$.
Jan
8
revised Is there a formula for the Frobenius-Schur indicator of a rep of a Lie group?
Added table.
Jan
8
answered Is there a formula for the Frobenius-Schur indicator of a rep of a Lie group?
Dec
24
reviewed Approve How to solve $f(f(x)) = \cos(x)$?
Dec
24
comment Manifolds as simultaneous coset spaces
The existence a $G$-equivariant map $F:X\to Y$, namely, one such that $F(gx)=gF(x)$ for all $g\in G$, $x\in X$, is a necessary and sufficient condition.
Oct
25
reviewed Approve Computational complexity of low rank SDP
Jul
27
revised Check symplectomorphism property on infinitesimal generators
Added explanation.
Jul
27
answered Check symplectomorphism property on infinitesimal generators
Jul
22
accepted Calculation with weights of $E_6$
Jun
9
awarded  Nice Answer
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16
awarded  Yearling
Feb
22
awarded  Nice Answer
Feb
21
comment Image of skew-symmetric bilinear map which is never zero on linearly independent vectors
$b:\Lambda^2\mathbb R^4\to W^3$ whose kernel is $a_{12}=a_{34}$, $a_{14}=a_{23}$, $a_{13}=-a_{24}$, and consider its projectivization $U^2 \subset \mathbb P (\Lambda^2\mathbb R^4)$. Then $U^2$ does not meet $\mathbb G(2,\mathbb R^4)$. On the other hand, $2\dim V - 3 = 5 > 3 = \dim W$.
Feb
21
comment Image of skew-symmetric bilinear map which is never zero on linearly independent vectors
the equation $a_{12}a_{34}-a_{13}a_{24}+a_{14}a_{23}=0$. Construct a linear map
Feb
21
comment Image of skew-symmetric bilinear map which is never zero on linearly independent vectors
I think this is a counter-example: $\mathbb G(2,\mathbb R^4)$ has dim $4$ and embedds in $\mathbb P(\Lambda^2\mathbb R^4)$ which has dim $5$, defined
Feb
21
comment Image of skew-symmetric bilinear map which is never zero on linearly independent vectors
This is way late, but I think the argument works over $\mathbb C$ but not over $\mathbb R$, which is what I wanted. The obvious idea of complexifying $B$ does not seem to preserve the assumption on $B$.
Jan
9
comment Image of skew-symmetric bilinear map which is never zero on linearly independent vectors
Simple and smart, thanks! In fact, the better bound suffices for what I need.
Jan
9
accepted Image of skew-symmetric bilinear map which is never zero on linearly independent vectors
Jan
9
asked Image of skew-symmetric bilinear map which is never zero on linearly independent vectors