bio | website | ime.usp.br/~gorodski |
---|---|---|
location | Sao Paulo, Brazil | |
age | 49 | |
visits | member for | 3 years, 10 months |
seen | yesterday | |
stats | profile views | 1,494 |
I am a Professor at the University of São Paulo. I have mostly worked on the interaction between Lie groups and geometry, more specifically, Lie transformation groups and submanifold geometry, but I am generally interested in different kinds of problems that can be considered from a geometrical viewpoint.
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 |
Jan 3 |
awarded | Nice Answer |
Oct 23 |
awarded | Necromancer |
Jul 20 |
comment |
Calculation with weights of $E_6$
@Jim: Thanks for your comment. I was sloppy: it was to mean $\Gamma$ is a subquotient of $W$. Anyway, I was wrong about the given motivation. Howe and Umeda's paper "The Capelli identity, the double commutant theorem, and multiplicity-free actions" discusses this representation. |
Jul 20 |
revised |
Calculation with weights of $E_6$
Updated information. |
Jul 17 |
comment |
Is the Duflo polynomial conjecture open?
I guess the main reference is: Journal of Functional Analysis, Volume 117, Issue 1, October 1993, Pages 174–214, Invariant Differential Operators in Symmetrical Spaces. II. Generalized Harish-Chandra Homomorphism, by C. Torossian. |
Jul 12 |
revised |
Calculation with weights of $E_6$
Added possible answer. |
Jul 11 |
revised |
Calculation with weights of $E_6$
Clarified notation. |
Jul 11 |
comment |
Calculation with weights of $E_6$
@Jim: I meant the Coxeter group of type $A_2$. |
Jul 11 |
comment |
Calculation with weights of $E_6$
@Vit: No, it is not. |
Jul 11 |
asked | Calculation with weights of $E_6$ |
Jul 2 |
awarded | Curious |
Jun 24 |
comment |
Is the group of isometries of a homogeneous Riemannian manifold maximal?
Of course. The odd-dimensional sphere $S^{2n+1}$ with the round metric is $O(2n+2)/O(2n+1)$. The subgroup $U(n+1)$ of $O(2n+2)$ acts transitively, and there are $U(n+1)$-invariant metrics which are not round. |