Difference between revisions of "Talk:Aufgaben:Problem 1"

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Ich verstehe diesen Schritt nicht ganz (Warum kommt das k da aus dem Integral raus?):
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The proof that \(\rho(e)=e\) is trivial, but I added it nevertheless. You will most likely lose points of you don't show it at the exam. It is very easy, though.
  
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\( \frac{1}{\sqrt{2\pi}}\int_{\mathbb{R}}-i\frac{k^2}{2}\hat g(k)e^{ikx-i\frac{k^2}{2}t}dk = -\frac{ik^2}{4\pi}\int_{\mathbb{R}}\hat g(k)e^{ikx-i\frac{k^2}{2}t}dk \)
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[[User:Lilit|Lilit]] ([[User talk:Lilit|talk]]) 12:37, 30 June 2015 (CEST)
  
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Grundsätzlich scheinen mir da einige Schritte noch nicht ganz sauber. Da fehlen einige limes und bei den Differenzenquotienten einige divisionen durch n. Zudem verstehe ich so nicht ganz warum man jetzt unter dem Integral differenzieren darf. Folgt dies jetzt mit dominierender Konvergenz, oder wie?
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I dont think it is necessary tho show that \(\rho\) preserves inverses for \(\rho\) to be a homomorphism, but it is certainly not wrong if you do it ;-)
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Die Differenzenquotienten verschwinden dann auch auf einmal. Aber vieleicht verstehe ich es einfach nicht ganz richtig.
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"[[User:Benjamin Kuhn|Benjamin Kuhn]] ([[User talk:Benjamin Kuhn|talk]]) 11:15, 2 January 2015 (CET)"
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[[User:Simfeld|Simfeld]] ([[User talk:Simfeld|talk]]) 11:57, 15 June 2015 (CEST)
  
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I assume you want to show that \(\rho(g) \in GL(V)\) but I believe this is always implied by: G a group, V a Vector space + homomorphism. So I also don't think it is necessary.
  
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Meiner Meinung nach müsste der Beweis etwa wo aussehen:
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[[User:Carl|Carl]] ([[User talk:Carl|talk]]) 19:15, 15 June 2015 (CEST)
  
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$$
 
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\frac{\partial}{\partial t} \int_{\mathbb{R}}\hat g(k)e^{ikx-i\frac{k^2}{2}t}dk \\
 
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= \lim\limits_{h \rightarrow 0} \int_{\mathbb{R}} \frac{1}{h}\hat g(k)e^{ikx}e^{-i\frac{k^2}{2}t}(e^{-i\frac{k^2}{2}h}-1)dk
 
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$$
 
  
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und jetzt unter dem integral abschätzen, damit man dominierende konvergenz anwenden kann
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Quote [https://en.wikipedia.org/wiki/Group_representation Wikipedia]:
  
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$$
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:: In [[mathematics]], given two [[group (mathematics)|groups]], (''G'', ∗) and (''H'', ·), a '''group homomorphism''' from (''G'', ∗) to (''H'', ·) is a [[function (mathematics)|function]] ''h'' : ''G'' → ''H'' such that for all ''u'' and ''v'' in ''G'' it holds that
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\lvert \frac{1}{h}\hat g(k)e^{ikx}e^{-i\frac{k^2}{2}t}(e^{-i\frac{k^2}{2}h}-1) \rvert \\
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::
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\le \lvert \frac{1}{h}\hat g(k) 2i \sin(\frac{k^2}{4}h) \rvert \\
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::: \( h(u*v) = h(u) \cdot h(v) \)
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\le \lvert \frac{1}{h}\hat g(k) 2 \frac{k^2}{4}h \rvert \\
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::
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\le \lvert \hat g(k) \frac{k^2}{2} \rvert
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:: where the group operation on the left hand side of the equation is that of ''G'' and on the right hand side that of ''H''.
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$$
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::
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and \( \hat g(k) \frac{k^2}{2} \in L^1 \)
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:: From this property, one can deduce that ''h'' maps the [[identity element]] ''e<sub>G</sub>'' of ''G'' to the identity element ''e<sub>H</sub>'' of ''H'', and it also maps inverses to inverses in the sense that
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::
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::: \( h\left(u^{-1}\right) = h(u)^{-1}. \,\)
  
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Damit haben wir die Bedinungen für dominierende Konvergenz erfüllt und können den Limes vom Differenzenquotienten unter das Integral ziehen, womit man unter dem Integral die Ableitung erhält.
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Doesn't that mean that we're already done after showing the homomorphism property?
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Also, writing \(\rho(e) = e\) seems wrong to me, as it's not the same entity. One \(e\) is in \(G\), the other should be in \(GL(V)\).
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--[[User:Nik|Nik]] ([[User talk:Nik|talk]]) 11:39, 26 July 2015 (CEST)
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But we don't have a group homomorphism until we have shown that the map is from \(G\) to "\(L(V)\)" is actually into \(GL(V)\) thus into a group. Am I wrong? It is pretty trial, as it always follows for any multiplicative map from a group to the linear maps on a Vector space.
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[[User:Carl|Carl]] ([[User talk:Carl|talk]]) 13:37, 26 July 2015 (CEST)
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I guess you're right. Thanks for clarifying!
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--[[User:Nik|Nik]] ([[User talk:Nik|talk]]) 15:45, 26 July 2015 (CEST)

Latest revision as of 13:45, 26 July 2015

The proof that \(\rho(e)=e\) is trivial, but I added it nevertheless. You will most likely lose points of you don't show it at the exam. It is very easy, though.

Lilit (talk) 12:37, 30 June 2015 (CEST)

I dont think it is necessary tho show that \(\rho\) preserves inverses for \(\rho\) to be a homomorphism, but it is certainly not wrong if you do it ;-)

Simfeld (talk) 11:57, 15 June 2015 (CEST)

I assume you want to show that \(\rho(g) \in GL(V)\) but I believe this is always implied by: G a group, V a Vector space + homomorphism. So I also don't think it is necessary.

Carl (talk) 19:15, 15 June 2015 (CEST)


Quote Wikipedia:

In mathematics, given two groups, (G, ∗) and (H, ·), a group homomorphism from (G, ∗) to (H, ·) is a function h : G → H such that for all u and v in G it holds that
\( h(u*v) = h(u) \cdot h(v) \)
where the group operation on the left hand side of the equation is that of G and on the right hand side that of H.
From this property, one can deduce that h maps the identity element eG of G to the identity element eH of H, and it also maps inverses to inverses in the sense that
\( h\left(u^{-1}\right) = h(u)^{-1}. \,\)

Doesn't that mean that we're already done after showing the homomorphism property?

Also, writing \(\rho(e) = e\) seems wrong to me, as it's not the same entity. One \(e\) is in \(G\), the other should be in \(GL(V)\).

--Nik (talk) 11:39, 26 July 2015 (CEST)

But we don't have a group homomorphism until we have shown that the map is from \(G\) to "\(L(V)\)" is actually into \(GL(V)\) thus into a group. Am I wrong? It is pretty trial, as it always follows for any multiplicative map from a group to the linear maps on a Vector space.

Carl (talk) 13:37, 26 July 2015 (CEST)

I guess you're right. Thanks for clarifying!

--Nik (talk) 15:45, 26 July 2015 (CEST)

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