Myprofileissaved

I found a vector of weight $(\mu_1, \mu_2)$ but am not sure how to proceed. I know every representation of $\mathfrak{sp}_4(\mathbb{C})$ is totally reducible.

Let $\mu_1 >= \mu_2$ be two natural numbers. Let $S = \mathbb{C}^4$ and let $A = \mathfrak{sp}(4,\mathbb{C})$ the complex symplectic Lie algebra of $Sp(4, \mathbb{C})$. Let $V$ be an irreducible representation of $A$ of highest weight (1,1). I need to prove $S^{\otimes(\mu_1 - \mu_2)}\otimes V^{\otimes \mu_2}$ has a irreducible sub-representation of weight (mu_1, mu_2).

Can someone give me a hint how to compute $[M_{g} ,S^{2}]$ where $M_{g}$ is the surface of genus g and $[M_{g} ,S^{2}]$ is the set of continuous maps from $M_{g}$ to the sphere modulo (free) homotopy ? My hint is to use the Cellular Approximation theorem. I know both are CW complexes.

$L:H^{1}( \Gamma (TM) ) \subset L^{2}( \Gamma (TM)) \rightarrow L^{2}( \Gamma (T^{\ast}M \otimes T^{\ast}M)) $

Lie derivative of the metric

how can I prove $L(x)=L_{x}g$ is a closed operator?

Is the product of two chain complexes (modules over ring R) is the product level wise and boundary maps points wise?

This makes no sense.

What does it mean that the functor from Set to Grp sending a set to the free group generated by it commutes with pushouts? what do I need to show? that given a pushout (P,i,j) with i,j the pushout morphisms, we have F(i)F(j)=F(j)F(i)?

Sorry, I mean $\langle\nabla u,\nabla u^{T}\rangle_{L^{2}(\Omega)}\geq0$

with $\Omega$ bounded

Hi, is it true that for $u\in C_{0}^{\infty}(\Omega)$ we have $$\langle\nabla u,\nabla u^{T}\rangle\geq0$$ ? if so, how can I show it?

smooth vector fields

$\mathfrak{X} (M)$ - domain of $\mathcal{L}$

Is it possible to show that with such norms $\mathcal{L}: \mathfrak(M) \rightarrow \Gamma(T^2 T^\ast M)$ taking $X\mapsto \mathcal{L}_{X}g$ is closed?

* sup going also over all $p\in M$

with $\Gamma(T^2 T^\ast M)$ being the space of covariant 2-tensor fields?

Ok. So for $T\in \Gamma{T^2 T^\ast M}$ can I take $Sup|T(X_1,X_2)|$ going over all $|X_i|=1$ $i=1,2$ as a norm? does that makes sense?

Yes, ofcourse.

$max_p |\langle X_p , X_p\rangle_{g_p}|^{1/2}$

Does it makes sense to take $Sup_p \langle X_p , X_p \rangle_{g_p}$ as a norm over vector fields?

&g& is the metric of the Riemannian manifold

Hi, I am trying to prove that the operator $X\mapsto \mathcal{L}_{X}g$ is a closed operator when working on a compact manifold. I know I need to limit my domain so that I will have a Norm space into a norm space, since the space of smooth sections over a manifold is not a norm space. Is there a way to do so?

I am mainly interested in the proof of one of the sides of the lemma (I am trying to proof $dimKerD < \infty$ )

Hi, I am looking for the source of the lemma in this question : math.stackexchange.com/questions/1594788/…

If A is a d by d matrix, how can I get this inequality $$|AA^{T}A-A|\leq C(1+|A|^3)$$ with the norm being any norm on $$\mathbb{R}^{d\times d}$$ ?

Thank you @TedShifrin

Oh this is cool!

or flat surface?

Ain't that meaning that my surface is like a sphere or ... not sure what is the opposite of a sphere

Ok. So far mean curvature is not in the material. But we are half way to the course

Yeah, the boundary is measurable

Yes. Lipschitz domain

Variational calculus

I don't know why I feel this is true but I got this ichy ich that I need to be careful when saying it :P

When $$ \Omega $$ is a bounded open connected set?

Is it true to say that the volume of $$\Omega \subset \mathbb{R}^n $$ only depends on $$ \partial \Omega $$ ?

Can someone assist me with math.stackexchange.com/questions/1462020/… , I am stuck in the same place and can't understand the equality

peace out

Order of magnitude more useful

without the need to restart the browser. Really nice

Yes it does

$f$ testing chatjax

Yes I understand now. Thank you very much robjohn!

oh ok. Now I understand :D

Yes. If $\eta(x-y)$ increasing in $x$ then it is decreasing in $y$.

I don't understand why the chain rule gives this

Why this equality holds? I don't get it

No. This $\int_{\Omega}\partial_{x}^{\alpha}\eta_{\delta}(x-y)f(y)dt=(-1)^{|a|}\int_{\Omega}\partial_{y}^{\alpha}\eta(x-y)f(y)dy$

ok. Any better way to present the question?