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

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Noch 3 Dinge, aber ansonsten sollte es ok sein:
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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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1)
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[[User:Lilit|Lilit]] ([[User talk:Lilit|talk]]) 12:37, 30 June 2015 (CEST)
  
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\(\check{\hat g}(x) = g(x)\)
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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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hier würde ich einfach erwähnen dass dies gilt, da \( g \in S(\mathbb{R}) \)
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[[User:Simfeld|Simfeld]] ([[User talk:Simfeld|talk]]) 11:57, 15 June 2015 (CEST)
  
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2)
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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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$$ \left| \frac{1}{s_n} \hat g (k) e^{ikx} e^{-i\frac{k^2}{2}t} \left( e^{i\frac{k^2}{2}s_n} -1 \right) \right|$$
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[[User:Carl|Carl]] ([[User talk:Carl|talk]]) 19:15, 15 June 2015 (CEST)
  
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Das müsste meiner Meinung nach um ein Vorzeichen korrigiert werden. Also:
 
  
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$$ \left| \frac{1}{s_n} \hat g (k) e^{ikx} e^{-i\frac{k^2}{2}t} \left( e^{-i\frac{k^2}{2}s_n} -1 \right) \right|$$
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Quote [https://en.wikipedia.org/wiki/Group_representation Wikipedia]:
  
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3)
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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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::: \( h(u*v) = h(u) \cdot h(v) \)
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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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:: 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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Für die zweite Ableitung nach x wäre mein Vorschlag:
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Doesn't that mean that we're already done after showing the homomorphism property?
  
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We can repeat the estimate we did for the first derivative and obviously it results in
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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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$$ \left| \frac{1}{s_n} \hat k g (k) e^{ikx} e^{-i\frac{k^2}{2}t} \left( e^{iks_n} -1 \right) \right| \leq \left| k^2 \cdot \hat g(k) \right| \in \mathcal{S} ({\mathbb{R}}) \subset L^1(\mathbb{R} ) $$
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--[[User:Nik|Nik]] ([[User talk:Nik|talk]]) 11:39, 26 July 2015 (CEST)
  
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Thus we use again dominated convergence and get the second derivative inside.
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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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