12,332 reputation
1034
bio website mathsci.appstate.edu/~cookwj
location Boone, NC
age 34
visits member for 2 years, 11 months
seen 4 hours ago

Assist Prof. at Appalachian State University (in Western North Carolina)


10h
comment On the construction of the Verma module
The answer is a little involved. But what you are trying to fight through is exact Proposition 10.5 (page 180) in Roger Carter's "Lie Algebras of Finite and Affine Type". If you don't have that text, I would highly recommend getting a copy. It's a truly excellent book. :)
12h
comment Prove Two Topologies Equivalent
Yes. There's nothing special about $\mathbb{R}^n$ here. Showing that "the subspace topology and the topology induced by restricting the metric to a subspace are equal" is exercise #1 in section 21 of Munkres' Topology. :)
22h
answered Prove Two Topologies Equivalent
1d
comment Rate Of Change Of Shadow
Looks fine to me.
1d
answered Weight spaces of Verma modules
1d
comment Verma modules and delta function
I suspect you aren't going to get a satisfactory answer here. You might want to try reposting on mathoverflow.net
Jul
2
awarded  Curious
Jun
27
awarded  Necromancer
Jun
22
comment How can a subspace have a lower dimension than its parent space?
Oh wait! I think I see what you're misunderstanding...the number of coordinates used to write down a vector isn't the same thing as the dimension of a subspace. I'll edit my post to try to help clear this up.
Jun
22
revised How can a subspace have a lower dimension than its parent space?
added 1306 characters in body
Jun
22
comment How can a subspace have a lower dimension than its parent space?
A subspace is a vector space sitting inside another vector space. It can be -- but does not have to be -- the whole thing. If we required that subspaces have dimensions matching their parent spaces, the concept of subspace would be pretty silly. Why? Any subspace whose dimension matches that of the parent space must in fact be equal to the whole parent space!
Jun
22
answered How can a subspace have a lower dimension than its parent space?
Jun
16
comment Double integral area.
One more thing...the first integral which you wrote down with order of integration: $dx\,dy$ is wrong for other reasons. You should never have a variable of integration appear in a bound for an integral outside where it appears as the variable of integration -- for a problem like this only numbers should appear as outermost bounds.
Jun
16
comment Double integral area.
Bounds should go "lower" to "upper". Notice that $y=2x^2-2$ is a parabola which opens upward and $y=-x^2+1$ is a parabola which opens downward. Thus $2x^2-2$ should be a lower bound and $1-x^2$ an upper bound.
Jun
16
comment Double integral area.
Your inner bounds are backwards. "$2x^2-2$" should be on the bottom and "$1-x^2$" on the top.
Jun
10
comment Does the Product of conjugates of some subgroups commutes with all elements of another subgroup
Glad to help! By the way, if you study group representation theory, you'll see this reindex "trick" used over and over again. For example, the above calculation looks a lot like what comes up when trying to determine the center of a group algebra.
Jun
10
revised Does the Product of conjugates of some subgroups commutes with all elements of another subgroup
added 491 characters in body
Jun
10
answered Does the Product of conjugates of some subgroups commutes with all elements of another subgroup
Jun
6
comment Finding vector orthogonal to two vectors
It should work. Gram-Schmidt works as long as the bilinear form (in particular for you "metric") is non-degenerate. It doesn't need to be positive definite to work. You should double check you calculation and make use used $w'$ in the final step not $w$.
Jun
6
comment Can someone what this notation means?
No. $k=$ the number of inputs $=$ the length of $I$, but this does not have to be the same as the dimension of $V$. One of the examples in my comment above has 3 inputs but they are drawn from a 2 dimensional vector space.