About representations of Lie groups

In summary: The representation theorem about unique irreducible representations of ##\mathfrak{sl}(2,\mathbb{F})## per dimension works for ##\mathbb{F}=\mathbb{R}## as well as far as I could see. The characteristic of ##\mathbb{F}## is crucial here, not the closure. So I don't know why the exercise uses complex numbers. Would be interesting to figure out. Maybe someone here can...
  • #1
aalma
46
1
Homework Statement
See below
Relevant Equations
We recall that the fundamental group of ##SL(2,R)## is ##Z## and define by ##G## the universal covering group for ##SL(2,R)##.
Show that any complex representation of G, ##ρ : G →GL(n,C)##, factors through ##SL(2,R)## (that is ##ρ## is the composition ##G →SL(2,R) →GL(n,C)##).

Hint: Any representation ##ρ## gives rise to a representation of the Lie algebra ##sl(2, R) → gl(n,C)##. This extends to ##sl(2,C) → gl(n,C)## that can be uniquely lifted to ##SL(2,C) → GL(n,C)##. The composition with ##G → SL(2,R) → SL(2,C)## should coincide with ##ρ##.
I am not quite sure of how this works, i.e. of what exactly I need to do with the hint.
Any explanantion would be helpful!
 
Physics news on Phys.org
  • #2
aalma said:
Homework Statement: See below
Relevant Equations: We recall that the fundamental group of ##SL(2,R)## is ##Z## and define by ##G## the universal covering group for ##SL(2,R)##.
Show that any complex representation of G, ##ρ : G →GL(n,C)##, factors through ##SL(2,R)## (that is ##ρ## is the composition ##G →SL(2,R) →GL(n,C)##).

Hint: Any representation ##ρ## gives rise to a representation of the Lie algebra ##sl(2, R) → gl(n,C)##. This extends to ##sl(2,C) → gl(n,C)## that can be uniquely lifted to ##SL(2,C) → GL(n,C)##. The composition with ##G → SL(2,R) → SL(2,C)## should coincide with ##ρ##.

I am not quite sure of how this works, i.e. of what exactly I need to do with the hint.
Any explanantion would be helpful!
I'm not sure whether this helps, but the representations of ##\mathfrak{sl}(2,\mathbb{C})## are all known: one irreducible representation (up to isomorphisms) per dimension ##n.##
 
  • Like
Likes aalma and topsquark
  • #3
fresh_42 said:
I'm not sure whether this helps, but the representations of ##\mathfrak{sl}(2,\mathbb{C})## are all known: one irreducible representation (up to isomorphisms) per dimension ##n.##
I cannot see how to connect this to my question
 
  • #4
aalma said:
I cannot see how to connect this to my question
We have a group homomorphism
\begin{align*}
\rho\, : \,G\longrightarrow GL(n,\mathbb{C})
\end{align*}
This results in a Lie algebra homomorphism
\begin{align*}
D\rho\, : \,\mathfrak{g}\longrightarrow \mathfrak{gl}(n,\mathbb{C})
\end{align*}
Now we need a Lie algebra homomorphism to put in between,
\begin{align*}
D\pi \, : \,\mathfrak{g} \longrightarrow \mathfrak{sl}(2,\mathbb{R})
\end{align*}
This looks as if it is done by the covering property. Use your definitions.
The extension to complex numbers is
\begin{align*}
\mathfrak{g} \stackrel{D\pi}{\longrightarrow } \mathfrak{sl}(2,\mathbb{R}) \longrightarrow \mathfrak{sl}(2,\mathbb{C})=\mathfrak{sl}(2,\mathbb{R})\otimes \mathbb{C}
\end{align*}
Next, we add an irreducible representation ##D\varphi ## of ##\mathfrak{sl}(2,\mathbb{C})## in ##\mathfrak{gl}(n,\mathbb{C})## which there is uniquely one per dimension. All together we get
\begin{align*}
D\rho \, : \,\mathfrak{g} \stackrel{D\pi}{\longrightarrow } \mathfrak{sl}(2,\mathbb{R})\longrightarrow \mathfrak{sl}(2,\mathbb{C})=\mathfrak{sl}(2,\mathbb{R})\otimes \mathbb{C}\stackrel{D\varphi }{\longrightarrow }\mathfrak{gl}(n,\mathbb{C})
\end{align*}
The lift (exponential function) gets us back into the world of groups with the ##SL(2,\mathbb{R})## in between as desired.
 
  • Like
Likes topsquark and aalma
  • #5
Thanks! I have some questions if it is okay.
*Does the "D" in ##D\pi## (for example) denote?
*Can tou please mention the statement of the covering property you're using here.
*Do we actually need to know here that ##sl_2(C)=sl_2(R)×C## (I am not sure if it something I can use or it is just like a direct sum)?
 
Last edited:
  • #6
The hint solves the question. Which part exactly is not clear to you?
 
  • Like
Likes topsquark and aalma
  • #7
martinbn said:
The hint solves the question. Which part exactly is not clear to you?
Yeah, I guess that too, but what am I supposed to add then?
They mention that "the fundamental group of ##SL(2,R)## is ##Z## and define by ##G## the universal covering group for ##SL(2,R)##", where do we exactly use this information?
##\rho:G\to GL(n,C)## so how it then gives rise to the rep mentioned in hint which is from ##sl(2,R)##..
 
Last edited:
  • #8
aalma said:
Yeah, I guess that too, but what am I supposed to add then?
May be it is meant to make you think about it and make sure you understand all the consepts involved. Where is it from?
aalma said:
They mention that "the fundamental group of ##SL(2,R)## is ##Z## and define by ##G## the universal covering group for ##SL(2,R)##", where do we exactly use this information?
##\rho:G\to GL(n,C)## so how it then gives rise to the rep mentioned in hint whuch is from ##sl(2,R)##..
The group and a covering group have the same algebra.
 
  • Like
Likes topsquark and aalma
  • #9
aalma said:
Thanks! I have some questions if it is okay.
*Does the "D" in ##D\pi## (for example) denote?
The hint speaks of the transition from the group level to the algebra level. This is the transition from an analytical manifold (groups) to their tangent spaces (Lie algebras). "D" denoted the derivative at the neutral element of the group homomorphisms involved. You can get it e.g. by considering paths in the groups through the identity element and differentiating them there.

This is of course my notation and you should use the one in your class. Since the hint suggests it, there has to be a notation for this.

aalma said:
*Can tou please mention the statement of the covering property you're using here.
See @martinbn 's post below, #8.
aalma said:
*Do we actually need to know here that ##sl_2(C)=sl_2(R)×C## (I am not sure if it something I can use or it is just like a direct sum)?
No, we don't. Since I know nothing about your context, I simply added how a complexification is usually done. It is not a direct sum, it is a tensor product that extends the scalar field.
https://en.wikipedia.org/wiki/Complexification
It is a bit more complicated on the group level
https://en.wikipedia.org/wiki/Complexification_(Lie_group)

The representation theorem about unique irreducible representations of ##\mathfrak{sl}(2,\mathbb{F})## per dimension works for ##\mathbb{F}=\mathbb{R}## as well as far as I could see. The characteristic of ##\mathbb{F}## is crucial here, not the closure. So I don't know why the exercise uses complex numbers. Would be interesting to figure out. Maybe someone here can help.
 
  • Like
Likes aalma and topsquark

Similar threads

  • Differential Geometry
Replies
1
Views
2K
  • Linear and Abstract Algebra
Replies
1
Views
642
Replies
27
Views
1K
Replies
9
Views
1K
Replies
5
Views
1K
  • Linear and Abstract Algebra
Replies
7
Views
1K
  • Calculus and Beyond Homework Help
Replies
6
Views
1K
  • Differential Equations
Replies
9
Views
3K
  • Calculus and Beyond Homework Help
Replies
6
Views
2K
  • High Energy, Nuclear, Particle Physics
Replies
27
Views
3K
Back
Top