Effect of a Lie group morphism associated to a Lie algebra morphism to wedge product
I currently struggeling with the last exercise on my assignment:
Fix $omegainbigwedge^3(mathbb R^n)^*$. Let $G$ be a Lie group, $rhocolon Gto GL(n,mathbb R)$ a Lie group morphism such that
$$
omega(rho(g)w_1, rho(g)w_2, rho(g)w_3) = omega(w_1,w_2,w_3)
$$
for all $gin G$ and $w_iinmathbb R^n$. Now we take the associated Lie algebra morphism $R = (rho)_{*e} colon mathfrak{g}to Mat(n,mathbb R)$.
Show that
$$
omega(R(v)w_1,w_2,w_3)+omega(w_1,R(v)w_2,w_3)+omega(w_1,w_2,R(v)w_3) = 0
$$
for all $vinmathfrak g$ and $w_iinmathbb R^n$.
It looks to me like somehow differentiating the first equation and ending up with the second but to apply the exterior derivative of a k-form we need another representation.
Another thought was using the relation
$$
omega(rho(g)w_1, rho(g)w_2, rho(g)w_3) = det(rho(g))omega(w_1,w_2,w_3)
$$
thus we see that $det(rho(g))=1$ by the first equation and we get $R(v)$ has trace zero.
Any help would be very much appreciated. :-)
differential-geometry lie-groups lie-algebras
asked Dec 26 '18 at 15:41
flaxflax
32
add a comment |
I currently struggeling with the last exercise on my assignment:
Fix $omegainbigwedge^3(mathbb R^n)^*$. Let $G$ be a Lie group, $rhocolon Gto GL(n,mathbb R)$ a Lie group morphism such that
$$
omega(rho(g)w_1, rho(g)w_2, rho(g)w_3) = omega(w_1,w_2,w_3)
$$
for all $gin G$ and $w_iinmathbb R^n$. Now we take the associated Lie algebra morphism $R = (rho)_{*e} colon mathfrak{g}to Mat(n,mathbb R)$.
Show that
$$
omega(R(v)w_1,w_2,w_3)+omega(w_1,R(v)w_2,w_3)+omega(w_1,w_2,R(v)w_3) = 0
$$
for all $vinmathfrak g$ and $w_iinmathbb R^n$.
It looks to me like somehow differentiating the first equation and ending up with the second but to apply the exterior derivative of a k-form we need another representation.
Another thought was using the relation
$$
omega(rho(g)w_1, rho(g)w_2, rho(g)w_3) = det(rho(g))omega(w_1,w_2,w_3)
$$
thus we see that $det(rho(g))=1$ by the first equation and we get $R(v)$ has trace zero.
Any help would be very much appreciated. :-)
differential-geometry lie-groups lie-algebras
asked Dec 26 '18 at 15:41
flaxflax
32
add a comment |
I currently struggeling with the last exercise on my assignment:
Fix $omegainbigwedge^3(mathbb R^n)^*$. Let $G$ be a Lie group, $rhocolon Gto GL(n,mathbb R)$ a Lie group morphism such that
$$
omega(rho(g)w_1, rho(g)w_2, rho(g)w_3) = omega(w_1,w_2,w_3)
$$
for all $gin G$ and $w_iinmathbb R^n$. Now we take the associated Lie algebra morphism $R = (rho)_{*e} colon mathfrak{g}to Mat(n,mathbb R)$.
Show that
$$
omega(R(v)w_1,w_2,w_3)+omega(w_1,R(v)w_2,w_3)+omega(w_1,w_2,R(v)w_3) = 0
$$
for all $vinmathfrak g$ and $w_iinmathbb R^n$.
It looks to me like somehow differentiating the first equation and ending up with the second but to apply the exterior derivative of a k-form we need another representation.
Another thought was using the relation
$$
omega(rho(g)w_1, rho(g)w_2, rho(g)w_3) = det(rho(g))omega(w_1,w_2,w_3)
$$
thus we see that $det(rho(g))=1$ by the first equation and we get $R(v)$ has trace zero.
Any help would be very much appreciated. :-)
differential-geometry lie-groups lie-algebras
asked Dec 26 '18 at 15:41
flaxflax
32
I currently struggeling with the last exercise on my assignment:
Fix $omegainbigwedge^3(mathbb R^n)^*$. Let $G$ be a Lie group, $rhocolon Gto GL(n,mathbb R)$ a Lie group morphism such that
$$
omega(rho(g)w_1, rho(g)w_2, rho(g)w_3) = omega(w_1,w_2,w_3)
$$
for all $gin G$ and $w_iinmathbb R^n$. Now we take the associated Lie algebra morphism $R = (rho)_{*e} colon mathfrak{g}to Mat(n,mathbb R)$.
Show that
$$
omega(R(v)w_1,w_2,w_3)+omega(w_1,R(v)w_2,w_3)+omega(w_1,w_2,R(v)w_3) = 0
$$
for all $vinmathfrak g$ and $w_iinmathbb R^n$.
It looks to me like somehow differentiating the first equation and ending up with the second but to apply the exterior derivative of a k-form we need another representation.
Another thought was using the relation
$$
omega(rho(g)w_1, rho(g)w_2, rho(g)w_3) = det(rho(g))omega(w_1,w_2,w_3)
$$
thus we see that $det(rho(g))=1$ by the first equation and we get $R(v)$ has trace zero.
Any help would be very much appreciated. :-)
differential-geometry lie-groups lie-algebras
differential-geometry lie-groups lie-algebras
asked Dec 26 '18 at 15:41
flaxflax
32
asked Dec 26 '18 at 15:41
flaxflax
32
asked Dec 26 '18 at 15:41
flaxflax
32
asked Dec 26 '18 at 15:41
flaxflax
32
asked Dec 26 '18 at 15:41
flaxflax
32
32
add a comment |
add a comment |
1 Answer
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If $vinmathfrak g$ and $tinmathbb R$, then$$omegabigl(rho(e^{tv})w_1,rho(e^{tv})w_2,rho(e^{tv})w_3bigr)=omega(w_1,w_2,w_3).$$Differentiating both sides at $t=0$, you get$$omegabigl(R(v)w_1,w_2,w_3bigr)+omegabigl(w_1,R(v)w_2,w_3bigr)+omegabigl(w_1,w_2,R(v)w_3bigr)=0.$$
answered Dec 26 '18 at 15:51
José Carlos SantosJosé Carlos Santos
152k22123225
This already helped a lot, thanks! But I have two little questions: For your first equation there has to be a $vinmathfrak g$ for every $gin G$ such that $$ rho(g) = R(e^{tv}). $$ I could see why there exist a matrix $Ainmathbb R^{ntimes n}$ for every $Bin GL(n,mathbb R)$ such that $B=e^{A}$ but why does it still hold for the morphisms? For the differentiation I need $$ frac{mathrm d}{mathrm dt}|_{t=0} R(mathrm e^{tv}) = R(frac{mathrm d}{mathrm dt}|_{t=0} mathrm e^{tv}). $$ I think there was a property which said that morphisms and the differentiation commute?
– flax
Dec 26 '18 at 17:39
Concerning your first question, all that I need (and all that I used) was that$$(forall vinmathfrak{g})(forall tinmathbb{R}):e^{iv}in G.$$Now use the fact that$$left.frac{mathrm d}{mathrm dt}right|_{t=0}e^{tv}=v$$together with the chain rule.
– José Carlos Santos
Dec 26 '18 at 19:21
I think my main problem was the $e^{tv}$-notion because I have just a theorem which says that there exists a curve with the same properties but everything works now. Thank you very much! :-)
– flax
Dec 27 '18 at 10:18
I'm glad I could help.
– José Carlos Santos
Dec 27 '18 at 10:23
Unless you think that I am asking too much, perhaps that you could upvote my answer.
– José Carlos Santos
Dec 27 '18 at 10:26
|
show 3 more comments
Your Answer
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1 Answer
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1 Answer
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If $vinmathfrak g$ and $tinmathbb R$, then$$omegabigl(rho(e^{tv})w_1,rho(e^{tv})w_2,rho(e^{tv})w_3bigr)=omega(w_1,w_2,w_3).$$Differentiating both sides at $t=0$, you get$$omegabigl(R(v)w_1,w_2,w_3bigr)+omegabigl(w_1,R(v)w_2,w_3bigr)+omegabigl(w_1,w_2,R(v)w_3bigr)=0.$$
answered Dec 26 '18 at 15:51
José Carlos SantosJosé Carlos Santos
152k22123225
This already helped a lot, thanks! But I have two little questions: For your first equation there has to be a $vinmathfrak g$ for every $gin G$ such that $$ rho(g) = R(e^{tv}). $$ I could see why there exist a matrix $Ainmathbb R^{ntimes n}$ for every $Bin GL(n,mathbb R)$ such that $B=e^{A}$ but why does it still hold for the morphisms? For the differentiation I need $$ frac{mathrm d}{mathrm dt}|_{t=0} R(mathrm e^{tv}) = R(frac{mathrm d}{mathrm dt}|_{t=0} mathrm e^{tv}). $$ I think there was a property which said that morphisms and the differentiation commute?
– flax
Dec 26 '18 at 17:39
Concerning your first question, all that I need (and all that I used) was that$$(forall vinmathfrak{g})(forall tinmathbb{R}):e^{iv}in G.$$Now use the fact that$$left.frac{mathrm d}{mathrm dt}right|_{t=0}e^{tv}=v$$together with the chain rule.
– José Carlos Santos
Dec 26 '18 at 19:21
I think my main problem was the $e^{tv}$-notion because I have just a theorem which says that there exists a curve with the same properties but everything works now. Thank you very much! :-)
– flax
Dec 27 '18 at 10:18
I'm glad I could help.
– José Carlos Santos
Dec 27 '18 at 10:23
Unless you think that I am asking too much, perhaps that you could upvote my answer.
– José Carlos Santos
Dec 27 '18 at 10:26
|
show 3 more comments
If $vinmathfrak g$ and $tinmathbb R$, then$$omegabigl(rho(e^{tv})w_1,rho(e^{tv})w_2,rho(e^{tv})w_3bigr)=omega(w_1,w_2,w_3).$$Differentiating both sides at $t=0$, you get$$omegabigl(R(v)w_1,w_2,w_3bigr)+omegabigl(w_1,R(v)w_2,w_3bigr)+omegabigl(w_1,w_2,R(v)w_3bigr)=0.$$
answered Dec 26 '18 at 15:51
José Carlos SantosJosé Carlos Santos
152k22123225
This already helped a lot, thanks! But I have two little questions: For your first equation there has to be a $vinmathfrak g$ for every $gin G$ such that $$ rho(g) = R(e^{tv}). $$ I could see why there exist a matrix $Ainmathbb R^{ntimes n}$ for every $Bin GL(n,mathbb R)$ such that $B=e^{A}$ but why does it still hold for the morphisms? For the differentiation I need $$ frac{mathrm d}{mathrm dt}|_{t=0} R(mathrm e^{tv}) = R(frac{mathrm d}{mathrm dt}|_{t=0} mathrm e^{tv}). $$ I think there was a property which said that morphisms and the differentiation commute?
– flax
Dec 26 '18 at 17:39
Concerning your first question, all that I need (and all that I used) was that$$(forall vinmathfrak{g})(forall tinmathbb{R}):e^{iv}in G.$$Now use the fact that$$left.frac{mathrm d}{mathrm dt}right|_{t=0}e^{tv}=v$$together with the chain rule.
– José Carlos Santos
Dec 26 '18 at 19:21
I think my main problem was the $e^{tv}$-notion because I have just a theorem which says that there exists a curve with the same properties but everything works now. Thank you very much! :-)
– flax
Dec 27 '18 at 10:18
I'm glad I could help.
– José Carlos Santos
Dec 27 '18 at 10:23
Unless you think that I am asking too much, perhaps that you could upvote my answer.
– José Carlos Santos
Dec 27 '18 at 10:26
|
show 3 more comments
If $vinmathfrak g$ and $tinmathbb R$, then$$omegabigl(rho(e^{tv})w_1,rho(e^{tv})w_2,rho(e^{tv})w_3bigr)=omega(w_1,w_2,w_3).$$Differentiating both sides at $t=0$, you get$$omegabigl(R(v)w_1,w_2,w_3bigr)+omegabigl(w_1,R(v)w_2,w_3bigr)+omegabigl(w_1,w_2,R(v)w_3bigr)=0.$$
answered Dec 26 '18 at 15:51
José Carlos SantosJosé Carlos Santos
152k22123225
If $vinmathfrak g$ and $tinmathbb R$, then$$omegabigl(rho(e^{tv})w_1,rho(e^{tv})w_2,rho(e^{tv})w_3bigr)=omega(w_1,w_2,w_3).$$Differentiating both sides at $t=0$, you get$$omegabigl(R(v)w_1,w_2,w_3bigr)+omegabigl(w_1,R(v)w_2,w_3bigr)+omegabigl(w_1,w_2,R(v)w_3bigr)=0.$$
answered Dec 26 '18 at 15:51
José Carlos SantosJosé Carlos Santos
152k22123225
edited Jan 4 at 14:30
answered Dec 26 '18 at 15:51
José Carlos SantosJosé Carlos Santos
152k22123225
answered Dec 26 '18 at 15:51
José Carlos SantosJosé Carlos Santos
152k22123225
answered Dec 26 '18 at 15:51
José Carlos SantosJosé Carlos Santos
152k22123225
152k22123225
This already helped a lot, thanks! But I have two little questions: For your first equation there has to be a $vinmathfrak g$ for every $gin G$ such that $$ rho(g) = R(e^{tv}). $$ I could see why there exist a matrix $Ainmathbb R^{ntimes n}$ for every $Bin GL(n,mathbb R)$ such that $B=e^{A}$ but why does it still hold for the morphisms? For the differentiation I need $$ frac{mathrm d}{mathrm dt}|_{t=0} R(mathrm e^{tv}) = R(frac{mathrm d}{mathrm dt}|_{t=0} mathrm e^{tv}). $$ I think there was a property which said that morphisms and the differentiation commute?
– flax
Dec 26 '18 at 17:39
Concerning your first question, all that I need (and all that I used) was that$$(forall vinmathfrak{g})(forall tinmathbb{R}):e^{iv}in G.$$Now use the fact that$$left.frac{mathrm d}{mathrm dt}right|_{t=0}e^{tv}=v$$together with the chain rule.
– José Carlos Santos
Dec 26 '18 at 19:21
I think my main problem was the $e^{tv}$-notion because I have just a theorem which says that there exists a curve with the same properties but everything works now. Thank you very much! :-)
– flax
Dec 27 '18 at 10:18
I'm glad I could help.
– José Carlos Santos
Dec 27 '18 at 10:23
Unless you think that I am asking too much, perhaps that you could upvote my answer.
– José Carlos Santos
Dec 27 '18 at 10:26
|
show 3 more comments
This already helped a lot, thanks! But I have two little questions: For your first equation there has to be a $vinmathfrak g$ for every $gin G$ such that $$ rho(g) = R(e^{tv}). $$ I could see why there exist a matrix $Ainmathbb R^{ntimes n}$ for every $Bin GL(n,mathbb R)$ such that $B=e^{A}$ but why does it still hold for the morphisms? For the differentiation I need $$ frac{mathrm d}{mathrm dt}|_{t=0} R(mathrm e^{tv}) = R(frac{mathrm d}{mathrm dt}|_{t=0} mathrm e^{tv}). $$ I think there was a property which said that morphisms and the differentiation commute?
– flax
Dec 26 '18 at 17:39
Concerning your first question, all that I need (and all that I used) was that$$(forall vinmathfrak{g})(forall tinmathbb{R}):e^{iv}in G.$$Now use the fact that$$left.frac{mathrm d}{mathrm dt}right|_{t=0}e^{tv}=v$$together with the chain rule.
– José Carlos Santos
Dec 26 '18 at 19:21
I think my main problem was the $e^{tv}$-notion because I have just a theorem which says that there exists a curve with the same properties but everything works now. Thank you very much! :-)
– flax
Dec 27 '18 at 10:18
I'm glad I could help.
– José Carlos Santos
Dec 27 '18 at 10:23
Unless you think that I am asking too much, perhaps that you could upvote my answer.
– José Carlos Santos
Dec 27 '18 at 10:26
This already helped a lot, thanks! But I have two little questions: For your first equation there has to be a $vinmathfrak g$ for every $gin G$ such that $$ rho(g) = R(e^{tv}). $$ I could see why there exist a matrix $Ainmathbb R^{ntimes n}$ for every $Bin GL(n,mathbb R)$ such that $B=e^{A}$ but why does it still hold for the morphisms? For the differentiation I need $$ frac{mathrm d}{mathrm dt}|_{t=0} R(mathrm e^{tv}) = R(frac{mathrm d}{mathrm dt}|_{t=0} mathrm e^{tv}). $$ I think there was a property which said that morphisms and the differentiation commute?
– flax
Dec 26 '18 at 17:39
This already helped a lot, thanks! But I have two little questions: For your first equation there has to be a $vinmathfrak g$ for every $gin G$ such that $$ rho(g) = R(e^{tv}). $$ I could see why there exist a matrix $Ainmathbb R^{ntimes n}$ for every $Bin GL(n,mathbb R)$ such that $B=e^{A}$ but why does it still hold for the morphisms? For the differentiation I need $$ frac{mathrm d}{mathrm dt}|_{t=0} R(mathrm e^{tv}) = R(frac{mathrm d}{mathrm dt}|_{t=0} mathrm e^{tv}). $$ I think there was a property which said that morphisms and the differentiation commute?
– flax
Dec 26 '18 at 17:39
Concerning your first question, all that I need (and all that I used) was that$$(forall vinmathfrak{g})(forall tinmathbb{R}):e^{iv}in G.$$Now use the fact that$$left.frac{mathrm d}{mathrm dt}right|_{t=0}e^{tv}=v$$together with the chain rule.
– José Carlos Santos
Dec 26 '18 at 19:21
Concerning your first question, all that I need (and all that I used) was that$$(forall vinmathfrak{g})(forall tinmathbb{R}):e^{iv}in G.$$Now use the fact that$$left.frac{mathrm d}{mathrm dt}right|_{t=0}e^{tv}=v$$together with the chain rule.
– José Carlos Santos
Dec 26 '18 at 19:21
I think my main problem was the $e^{tv}$-notion because I have just a theorem which says that there exists a curve with the same properties but everything works now. Thank you very much! :-)
– flax
Dec 27 '18 at 10:18
I think my main problem was the $e^{tv}$-notion because I have just a theorem which says that there exists a curve with the same properties but everything works now. Thank you very much! :-)
– flax
Dec 27 '18 at 10:18
I'm glad I could help.
– José Carlos Santos
Dec 27 '18 at 10:23
I'm glad I could help.
– José Carlos Santos
Dec 27 '18 at 10:23
Unless you think that I am asking too much, perhaps that you could upvote my answer.
– José Carlos Santos
Dec 27 '18 at 10:26
Unless you think that I am asking too much, perhaps that you could upvote my answer.
– José Carlos Santos
Dec 27 '18 at 10:26
|
show 3 more comments
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