Difference between revisions of "Talk:Aufgaben:Problem 13"

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Consider the initial value problem (IVP)
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There is an alternative way of proving (a) in the Felder Script:
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$$
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\begin{align}
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i\frac{\partial}{\partial t} f(x,t) &=-\frac{1}{2} \frac{\partial^2}{\partial x^2} f(x,t) \\
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f(x,0) &= g(x) \in S(\mathbb{R}) \\ 
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\end{align}
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$$
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Show that
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$$
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\begin{align}
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f(x,t)=(\hat g(k)e^{-i\frac{k^2}{2}t})\check (x) \\
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\end{align}
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$$
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with  \(\check{\hat\phi}=\phi\) and \(\check\phi (x)=\frac{1}{\sqrt{2\pi}}\int_{\mathbb{R}}\phi(k)e^{ikx}dk\) '''''(1)''''', is a solution of (IVP).
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== Solution ==
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https://people.math.ethz.ch/~felder/mmp/mmp2/
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Rewrite the possible solution with help of '''''(1)''''':
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$$ \begin{align}
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f(x,t)=\left( \hat g(k) e^{-i\frac{k^2}{2}t} \right) \check (x) &=\frac{1}{\sqrt{2\pi}}\int_{\mathbb{R}}\hat{g(k)} e^{-i\frac{k^2}{2}t}e^{ikx} \, dk \\
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&= \frac{1}{\sqrt{2\pi}}\int_{\mathbb{R}}\hat g(k)e^{ikx-i\frac{k^2}{2}t} \, dk
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\end{align} $$
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Check if the solution satisfies the initial condition:
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see chapter 3 -> Satz 3.1. and chapter 2 -> Satz 2.6 for a prove of Schur's lemma.
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$$ f(x,0) = \frac{1}{\sqrt{2\pi}}\int_{\mathbb{R}}\hat g(k)e^{ikx-0}dk = \frac{1}{\sqrt{2\pi}}\int_{\mathbb{R}}\hat g(k)e^{ikx} \, dk =\check{\hat g}(x) = g(x) $$
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For this, we used again '''''(1)''''' and the identity  \(\check{\hat\phi}=\phi\).
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Check if the solution satisfies the differential equation: <br />
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It seems to be shorter, and thus is probably better for the exam... [[User:Carl|Carl]] ([[User talk:Carl|talk]]) 23:16, 13 June 2015 (CEST)
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Calculate de derivations:
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$$ \begin{align}
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\frac{\partial}{\partial t} f(x,t) &= \frac{\partial}{\partial t} \left( \frac{1}{\sqrt{2\pi}}\int_{\mathbb{R}}\hat g(k)e^{ikx-i\frac{k^2}{2}t}dk \right) \\
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&= \frac{1}{\sqrt{2\pi}}\int_{\mathbb{R}}\frac{\partial}{\partial t}\hat g(k)e^{ikx-i\frac{k^2}{2}t}dk \\
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&= \frac{1}{\sqrt{2\pi}}\int_{\mathbb{R}}-i\frac{k^2}{2}\hat g(k)e^{ikx-i\frac{k^2}{2}t}dk \\
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&= -\frac{i}{2\sqrt{2\pi}}\int_{\mathbb{R}}\hat g(k)e^{ikx-i\frac{k^2}{2}t}k^2dk
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\end{align} $$
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$$ \begin{align}
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----
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\frac{\partial^2}{\partial x^2} f(x,t) &= \frac{\partial^2}{\partial x^2} \left( \frac{1}{\sqrt{2\pi}}\int_{\mathbb{R}}\hat g(k)e^{ikx-i\frac{k^2}{2}t} \, dk \right) \\
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&= \frac{1}{\sqrt{2\pi}}\int_{\mathbb{R}}\frac{\partial^2}{\partial x^2}\hat g(k)e^{ikx-i\frac{k^2}{2}t} \, dk \\
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&= \frac{1}{\sqrt{2\pi}}\int_{\mathbb{R}}\left(i^2k^2 \right) \hat g(k)e^{ikx-i\frac{k^2}{2}t} \, dk \\
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&= -\frac{1}{\sqrt{2\pi}}\int_{\mathbb{R}}\hat g(k)e^{ikx-i\frac{k^2}{2}t}k^2 \, dk
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\end{align} $$
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Put the whole thing in the differential equation and we see:
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Did anyone succeed in getting a shorter solution for a) by using Schur's lemma (as suggested by Carl)?
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$$ i\frac{\partial}{\partial t} f(x,t) = -\frac{i^2}{2\sqrt{2\pi}}\int_{\mathbb{R}}\hat g(k)e^{ikx-i\frac{k^2}{2}t}k^2 \, dk =\frac{1}{2\sqrt{2\pi}}\int_{\mathbb{R}}\hat g(k)e^{ikx-i\frac{k^2}{2}t}k^2dk = \left( -\frac{1}{2}\right) \cdot \left( -\frac{1}{\sqrt{2\pi}}\int_{\mathbb{R}}\hat g(k)e^{ikx-i\frac{k^2}{2}t}k^2dk\right) = -\frac{1}{2}\frac{\partial^2}{\partial x^2} f(x,t) $$
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it is satisfied.
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''Proof of exchanging the derivation and the integral:''
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- [[User:Snick|Snick]] ([[User talk:Snick|talk]]) 13:06, 30 July 2015 (CEST)
  
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We want to apply the Lebesgue dominated convergence theorem. This is a standard argument being used in some other problems.  
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I don't think you can make it any shorter. Shur's Lemma is used implicitly in the proof (claim 2), and to it you
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must first show that T is a homomorphism of representations (i.e. claim 1). The Felder Skript provesClaim 2
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and Claim 5, but then you must still show the rest.
  
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First, define \( h(x,t,k)=\hat g(k)e^{ikx-i\frac{k^2}{2}t} \) and since \( g(x) \in  \mathcal{S} ({\mathbb{R}}) \rightarrow \hat g(k)  \in \mathcal{S} ({\mathbb{R}}) \), we get considering the following argument:
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- [[User:Rayan|Rayan]] ([[User talk:Rayan|talk]]) 11:36, 31 July 2015 (CEST)
  
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$$ \left| f(x) \right| \leq \frac{C}{(1 + \vert x \vert^2)} $$
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----
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Maybe a little shortening. Instead taking \(S^{(k,l)}\) and \(T^{(k,l)}\), take a more generall \(U^{(k,l),[\rho,\sigma]}_{ij}=(\rho_{ki},\sigma_{lj})_G\) for any two unitary representations \(\rho, \sigma \) and with \(U^{(k,l),[\rho,\sigma]}\sigma(g)=\rho(g) U^{(k,l),[\rho,\sigma]}\)  also \([T^{(k,l)},\rho(g)]=0\) is showed.
  
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for some \( C \in \mathbb{R} \) and thus:
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\(\sum\limits_{i,j}{U^{(i,j),[\rho,\sigma]}_{i,j}}=(ch(\rho),ch(\sigma))_G\) may also save some time
  
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$$ \int_{\mathbb{R}}  \left| f(x) \right| dx \leq \int_{\mathbb{R}}  \left| \frac{C}{(1 + \vert x \vert^2)} \right| dx < \infty $$
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--[[User:Brynerm|Brynerm]] ([[User talk:Brynerm|talk]]) 17:01, 4 August 2015 (CEST)
  
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that \( \hat g(k)  \in L^1 \).
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Very nice, that does same some time.
  
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We then want to consider
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[[User:Carl|Carl]] ([[User talk:Carl|talk]]) 17:21, 4 August 2015 (CEST)
  
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$$ \lim_{n \rightarrow \infty} \int_{\mathbb{R}} \frac{h(x + s_n, t, k) - h(x,t,k)}{s_n} \, dk $$
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----
  
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with \( s_n \) an arbitrary zero-sequence and see that  
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I get slightly confused, that you act \(L^{W}\) on \(\chi_{sy}\) although \(\chi_{sy}\) may not be in \(W\) (Claim 9). Does it come from the fact, that linearity of \(L^{W}\) is only assured in \(W\) and not in \(V_s\) so you would have to use \(L^{V_s}\) here and make the restriction afterwards?
  
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$$ \left| \frac{1}{s_n} \hat g (k) e^{ikx} e^{-i\frac{k^2}{2}t} \left( e^{iks_n} -1 \right) \right| = \left| \frac{1}{s_n} \hat g (k) \sin \left( \frac{ks_n}{2} \right)\right| \leq \left| \frac{2}{s_n} \hat g (k) \frac{ks_n}{2} \right| = \left| k \cdot \hat g(k) \right| \in \mathcal{S} ({\mathbb{R}}) \subset L^1(\mathbb{R} ) $$
 
  
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So \( \frac{d}{dx} \int_{\mathbb{R}} h(x,t,k) \, dk = \int_{\mathbb{R}} \frac{d}{dx} h(x,t,k) \, dk \), and since \( \frac{d}{dx} h(x,t,k) \) results in a Schwartz-function depending on \( k \) times \( h \) we can apply the Lebesgue-dominated-convergence theorem twice.
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--[[User:Brynerm|Brynerm]] ([[User talk:Brynerm|talk]]) 07:56, 5 August 2015 (CEST)
  
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It remains to be shown for \( \frac{d}{dt} \) which is basically just repeating the argument:
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No \(V_s\) is not always one-dimentional. This would mean that every irreducible rep. is one dimensional. \(V_s = W\) is only true for \(\text{dim}V_s = 1\)
  
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$$ \lim_{n \rightarrow \infty} \int_{\mathbb{R}} \frac{h(x, t  + s_n, k) - h(x,t,k)}{s_n} \, dk $$
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Your right that is not correct. I just changed the \(L^{V_s}\) to \(L^W\) in claim 9, without much thought, after they changed the exercise. Not sure if applying the restriction afterwards works. I guess you can bypass the step by just writing it out a bit more:
  
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again with \( s_n \) an arbitrary zero-sequence, and then we go on:
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$$\sum_{i= 1}^n (e_i, L^W(g) e_i)_G =\sum_{i= 1}^n \frac{1}{|G|}\sum_{h\in G} e_i(h) (L^W(g) e_i)(h)^* =\sum_{i= 1}^n \frac{1}{|G|}\sum_{h\in G} e_i(h) e_i(g^{-1}h)^*  = \sum_{i= 1}^n \frac{1}{|G|}\sum_{h\in G} \sum_{z\in G} \lambda_{iz} \chi_{sz}(h) \sum_{y\in G} \lambda_{iy}^* \chi_{sy}(g^{-1}h)^*$$
  
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$$ \left| \frac{1}{s_n} \hat g (k) e^{ikx} e^{-i\frac{k^2}{2}t} \left( e^{i\frac{k^2}{2}s_n} -1 \right) \right| = \left| \frac{1}{s_n} \hat g (k) \sin \left( \frac{k^2s_n}{4} \right)\right| \leq \left| \frac{2}{s_n} \hat g (k) \frac{k^2s_n}{4} \right| = \left| \frac{k^2}{2} \cdot \hat g(k) \right| \in \mathcal{S} ({\mathbb{R}}) \subset L^1(\mathbb{R} ) $$.
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$$= \sum_{i= 1}^n \frac{1}{|G|}\sum_{h\in G} \sum_{z\in G} \lambda_{iz} \chi_{sz}(h) \sum_{y\in G} \lambda_{iy}^* \chi_{sgy}(h)^* = \sum_{i= 1}^n \sum_{z\in G} \sum_{y\in G} \lambda_{iz} \lambda_{iy}^* (\chi_{sz}, \chi_{sgy} )_G = \dots$$
  
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''Ende der Vorstellung.''
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[[User:Carl|Carl]] ([[User talk:Carl|talk]]) 15:44, 5 August 2015 (CEST)
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----
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I messed that up with dimensions, I was thinking in terms of conjugacy classes, but that wouldn't imply W=Vs.
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Nevertheless your bypass seems quite accurate.
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--[[User:Brynerm|Brynerm]] ([[User talk:Brynerm|talk]]) 22:59, 5 August 2015 (CEST)

Latest revision as of 20:59, 5 August 2015

There is an alternative way of proving (a) in the Felder Script:

https://people.math.ethz.ch/~felder/mmp/mmp2/

see chapter 3 -> Satz 3.1. and chapter 2 -> Satz 2.6 for a prove of Schur's lemma.

It seems to be shorter, and thus is probably better for the exam... Carl (talk) 23:16, 13 June 2015 (CEST)

Did anyone succeed in getting a shorter solution for a) by using Schur's lemma (as suggested by Carl)?

- Snick (talk) 13:06, 30 July 2015 (CEST)

I don't think you can make it any shorter. Shur's Lemma is used implicitly in the proof (claim 2), and to it you must first show that T is a homomorphism of representations (i.e. claim 1). The Felder Skript provesClaim 2 and Claim 5, but then you must still show the rest.

- Rayan (talk) 11:36, 31 July 2015 (CEST)

Maybe a little shortening. Instead taking \(S^{(k,l)}\) and \(T^{(k,l)}\), take a more generall \(U^{(k,l),[\rho,\sigma]}_{ij}=(\rho_{ki},\sigma_{lj})_G\) for any two unitary representations \(\rho, \sigma \) and with \(U^{(k,l),[\rho,\sigma]}\sigma(g)=\rho(g) U^{(k,l),[\rho,\sigma]}\) also \([T^{(k,l)},\rho(g)]=0\) is showed.

\(\sum\limits_{i,j}{U^{(i,j),[\rho,\sigma]}_{i,j}}=(ch(\rho),ch(\sigma))_G\) may also save some time

--Brynerm (talk) 17:01, 4 August 2015 (CEST)

Very nice, that does same some time.

Carl (talk) 17:21, 4 August 2015 (CEST)

I get slightly confused, that you act \(L^{W}\) on \(\chi_{sy}\) although \(\chi_{sy}\) may not be in \(W\) (Claim 9). Does it come from the fact, that linearity of \(L^{W}\) is only assured in \(W\) and not in \(V_s\) so you would have to use \(L^{V_s}\) here and make the restriction afterwards?


--Brynerm (talk) 07:56, 5 August 2015 (CEST)

No \(V_s\) is not always one-dimentional. This would mean that every irreducible rep. is one dimensional. \(V_s = W\) is only true for \(\text{dim}V_s = 1\)

Your right that is not correct. I just changed the \(L^{V_s}\) to \(L^W\) in claim 9, without much thought, after they changed the exercise. Not sure if applying the restriction afterwards works. I guess you can bypass the step by just writing it out a bit more:

$$\sum_{i= 1}^n (e_i, L^W(g) e_i)_G =\sum_{i= 1}^n \frac{1}{|G|}\sum_{h\in G} e_i(h) (L^W(g) e_i)(h)^* =\sum_{i= 1}^n \frac{1}{|G|}\sum_{h\in G} e_i(h) e_i(g^{-1}h)^* = \sum_{i= 1}^n \frac{1}{|G|}\sum_{h\in G} \sum_{z\in G} \lambda_{iz} \chi_{sz}(h) \sum_{y\in G} \lambda_{iy}^* \chi_{sy}(g^{-1}h)^*$$

$$= \sum_{i= 1}^n \frac{1}{|G|}\sum_{h\in G} \sum_{z\in G} \lambda_{iz} \chi_{sz}(h) \sum_{y\in G} \lambda_{iy}^* \chi_{sgy}(h)^* = \sum_{i= 1}^n \sum_{z\in G} \sum_{y\in G} \lambda_{iz} \lambda_{iy}^* (\chi_{sz}, \chi_{sgy} )_G = \dots$$

Carl (talk) 15:44, 5 August 2015 (CEST)

I messed that up with dimensions, I was thinking in terms of conjugacy classes, but that wouldn't imply W=Vs.

Nevertheless your bypass seems quite accurate.

--Brynerm (talk) 22:59, 5 August 2015 (CEST)

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