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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$$
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\begin{align}
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\frac{\partial}{\partial t} f(x,t) &=-\frac{1}{2} \frac{\partial^2}{\partial x^2} f(x,t) \\
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f(x,0) &= g(x) \in S(\mathbb{R}) \\ 
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\end{align}
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$$
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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)=(\hat g(k)e^{-i\frac{k^2}{2}t})\check (x)&= \frac{1}{\sqrt{2\pi}}\int_{\mathbb{R}}\hat g(k)e^{-i\frac{k^2}{2}t}e^{ikx}dk &= \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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\begin{align}
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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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\end{align}
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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} (\frac{1}{\sqrt{2\pi}}\int_{\mathbb{R}}\hat g(k)e^{ikx-i\frac{k^2}{2}t}dk) &= \frac{1}{\sqrt{2\pi}}\int_{\mathbb{R}}\frac{\partial}{\partial t}\hat g(k)e^{ikx-i\frac{k^2}{2}t}dk &= \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 &= -\frac{1}{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} (\frac{1}{\sqrt{2\pi}}\int_{\mathbb{R}}\hat g(k)e^{ikx-i\frac{k^2}{2}t}dk) &= \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 &= \frac{1}{\sqrt{2\pi}}\int_{\mathbb{R}}i^2k^2\hat g(k)e^{ikx-i\frac{k^2}{2}t}dk &= -\frac{1}{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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Compare it with the differential equation:
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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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\begin{align}
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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^2dk &=\frac{1}{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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-\frac{1}{\sqrt{2\pi}}\frac{\partial^2}{\partial x^2} f(x,t)&= (-\frac{1}{2})*(-\frac{1}{\sqrt{2\pi}}\int_{\mathbb{R}}\hat g(k)e^{ikx-i\frac{k^2}{2}t}k^2dk) &= \frac{1}{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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It is the same. <br />
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== Prove for used exchange==
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- [[User:Snick|Snick]] ([[User talk:Snick|talk]]) 13:06, 30 July 2015 (CEST)
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Prove why we can exchange the derivation and the integral:
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$$ i\frac{\partial}{\partial t} \frac{1}{\sqrt{2\pi}} \int_{\mathbb{R}}\hat g(k)e^{ikx-i\frac{k^2}{2}t}dk=\frac{i}{\sqrt{2\pi}}\int_{\mathbb{R}}\frac{\partial}{\partial t}\hat g(k)e^{ikx-i\frac{k^2}{2}t}dk $$
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because
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$$ \lvert\frac{\partial}{\partial t}\hat g(k)e^{ikx-i\frac{k^2}{2}t}\rvert =  \lvert\lim\limits_{n \rightarrow 0}{\frac{1}{n}\hat g(k)e^{ikx}e^{-i\frac{k^2}{2}t}(e^{-i\frac{k^2}{2}n}-1)}\rvert = \frac{2}{n}\lvert\hat g(k)\rvert\lvert e^{-i\frac{k^2 x^2}{2}}\rvert\lvert \sin(\frac{k^2}{4})\rvert \underset{\overbrace{\lvert\sin(k^2)\rvert < \lvert k \rvert}}{\leqslant} \lvert k \rvert \lvert \hat g(k) \rvert \in \ L^1 (\text{since } g(k) \in S(\mathbb{R})\subset L^1)$$
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and
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$$  -\frac{1}{2}\frac{\partial^2}{\partial x^2}\frac{1}{\sqrt{2\pi}} \int_{\mathbb{R}}\hat g(k)e^{ikx-i\frac{k^2}{2}t}dk =  -\frac{1}{2}\frac{1}{\sqrt{2\pi}} \int_{\mathbb{R}}\frac{\partial^2}{\partial x^2}\hat g(k)e^{ikx-i\frac{k^2}{2}t}dk $$
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because
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$$ \lvert\frac{\partial}{\partial x}\hat g(k)e^{ikx-i\frac{k^2}{2}t}\rvert = \lvert\lim\limits_{n \rightarrow 0}{\frac{1}{n} \hat g(k)e^{ikx}e^{-i\frac{k^2}{2}t}(e^{ikn}-1)}\rvert = \frac{2}{n}\lvert\hat g(k)\rvert\lvert e^{i\frac{k^2 x^2}{2}}\rvert\lvert \sin(\frac{xkn}{2})\rvert \underset{\overbrace{\lvert\sin(x)\rvert \leqslant \lvert x \rvert}}{\leqslant} \lvert k \rvert \lvert \hat g(k) \rvert \underset{=1}{\underbrace{\lvert e^{ik^2 x^2} \rvert}} \in L^1 (\text{since } \hat g(k) \in S(\mathbb{R})\subset L^1) $$
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and since
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$$  \frac{\partial}{\partial x}\hat g(k) e^{-i\frac{k^2}{2}t + ikx} = k \hat g(k) e^{-i\frac{k^2}{2}t} $$
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and for \( p \in \mathbb{R}[x] \text{ and } f \in S(\mathbb{R}) \text{ and } p\circ f \in S(\mathbb{R}) \) we are also able to exchange the second derivative.
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Therefore we get
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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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$$ i \frac{\partial}{\partial t} \frac{1}{\sqrt{2\pi}} \int_{\mathbb{R}}\hat g(k)e^{ikx-i\frac{k^2}{2}t}dk = \frac{i}{\sqrt{2\pi}}\int_{\mathbb{R}}\frac{\partial}{\partial t}\hat g(k)e^{ikx-i\frac{k^2}{2}t}dk = -\frac{(i)^2}{2\sqrt{2\pi}}\int_{\mathbb{R}}\hat g(k)k^2 e^{ikx-i\frac{k^2}{2}t}dx =  \boldsymbol{(1)} $$
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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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$$  -\frac{1}{2}\frac{1}{\sqrt{2\pi}}\frac{\partial^2}{\partial x^2} \int_{\mathbb{R}}\hat g(k)e^{ikx-i\frac{k^2}{2}t}dx =  -\frac{1}{2} \frac{1}{\sqrt{2\pi}} \int_{\mathbb{R}}\frac{\partial^2}{\partial x^2} \hat g(k)e^{ikx-i\frac{k^2}{2}t}dk = -\frac{(i)^2}{2\sqrt{2\pi}}\int_{\mathbb{R}} g(k)k^2 e^{ikx-i\frac{k^2}{2}t}dk =  \boldsymbol{(2)} $$
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- [[User:Rayan|Rayan]] ([[User talk:Rayan|talk]]) 11:36, 31 July 2015 (CEST)
  
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$$  \boldsymbol{(1)} = \boldsymbol{(2)}  $$
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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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\(\sum\limits_{i,j}{U^{(i,j),[\rho,\sigma]}_{i,j}}=(ch(\rho),ch(\sigma))_G\) may also save some time
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 +
--[[User:Brynerm|Brynerm]] ([[User talk:Brynerm|talk]]) 17:01, 4 August 2015 (CEST)
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Very nice, that does same some time.
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[[User:Carl|Carl]] ([[User talk:Carl|talk]]) 17:21, 4 August 2015 (CEST)
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 +
----
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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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 +
 
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--[[User:Brynerm|Brynerm]] ([[User talk:Brynerm|talk]]) 07:56, 5 August 2015 (CEST)
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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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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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$$\sum_{i= 1}^n (e_i, L^W(g) e_i)_G =\sum_{i= 1}^n \frac{1}{|G|}\sum_{h\in G} e_i(h) (L^W(g) e_i)(h)^* =\sum_{i= 1}^n \frac{1}{|G|}\sum_{h\in G} e_i(h) e_i(g^{-1}h)^*  = \sum_{i= 1}^n \frac{1}{|G|}\sum_{h\in G} \sum_{z\in G} \lambda_{iz} \chi_{sz}(h) \sum_{y\in G} \lambda_{iy}^* \chi_{sy}(g^{-1}h)^*$$
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$$= \sum_{i= 1}^n \frac{1}{|G|}\sum_{h\in G} \sum_{z\in G} \lambda_{iz} \chi_{sz}(h) \sum_{y\in G} \lambda_{iy}^* \chi_{sgy}(h)^* = \sum_{i= 1}^n \sum_{z\in G}  \sum_{y\in G} \lambda_{iz} \lambda_{iy}^* (\chi_{sz}, \chi_{sgy} )_G = \dots$$
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[[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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