Further questions on the basics of Lie Algebras

vertices
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I'm reviewing my Group theory notes at the moment. I have a few questions.

1)what is the "connected component of the identity"? (How would you go about working this out?)

2)Why is Lie(SO(n))=Lie(O(n))?

3)Why is the SU(2) Lie group isomorphic to a sphere in 3D?
 
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vertices said:
1)what is the "connected component of the identity"? (How would you go about working this out?)
The largest subgroup that's also a connected topological space. I think that for Lie groups, "connected" is equivalent to "path connected", which means that what you have to check is that for each g in the group, there's a continuous curve from the identity to g.

vertices said:
2)Why is Lie(SO(n))=Lie(O(n))?
For matrix Lie groups, we can use the simple definition of a Lie algebra: A matrix X is in the Lie algebra if exp(tX) is in the Lie group for all real numbers t. This condition implies that for both Lie groups, the Lie algebra is the real vector space of traceless hermitian n×n matrices.

Note that since det(exp(tA))=exp(tTr(A)), both the conditions ±1=det(exp(tA) and +1=det(exp(tA)) imply the same thing, that Tr(A)=0.

This book can tell you the details.

vertices said:
3)Why is the SU(2) Lie group isomorphic to a sphere in 3D?
Not Lie group isomorphic. Just homeomorpic (topological space isomorphic) and probably diffeomorphic (manifold isomorphic). What you're supposed to prove is that there's a continuous bijection from SU(2) into S3, and this is just a tedius calculation. Write down the most general complex 2×2 matrix U and find out what relationships between its components you can derive from the conditions U^\dagger U=I and \det U=1. You should end up with the condition that defines the unit 3-sphere.
 
Thanks Fredrik for your help in this and other threads. Hope you had a good Christmas:)
 
Fredrik said:
Not Lie group isomorphic. Just homeomorpic (topological space isomorphic) and probably diffeomorphic (manifold isomorphic). What you're supposed to prove is that there's a continuous bijection from SU(2) into S3, and this is just a tedius calculation. Write down the most general complex 2×2 matrix U and find out what relationships between its components you can derive from the conditions U^\dagger U=I and \det U=1. You should end up with the condition that defines the unit 3-sphere.

I attempted this computation as you suggested. I get matrix elements that look like this (sorry I don't know how to do matrices in Latex):

a b
-b* a

So there are two parameters, a and b. But because a=x+iy and b=w+iz, that's 4 parameter in total. So should it not belong to a 4-sphere?
 
The fact that the determinant of the matrix must be =1 gives you w^2+x^2+y^2+z^2=1. That's the definition of a 3-sphere. Note that a 2-sphere is a sphere and a 1-sphere is a circle. The number indicates how many dimensions the manifold has. A circle is a 1-dimensional manifold because its coordinate systems are maps into \mathbb R, not into \mathbb R^2.

Matrices start with \begin{pmatrix} and end with \end{pmatrix}. End each line except the last with \\, and put & symbols between the elements. For example, \begin{pmatrix} a & b\\ -b^* & a\end{pmatrix}

\begin{pmatrix} a & b\\ -b^* & a\end{pmatrix}

Use vmatrix instead of pmatrix if you want a determinant.
 

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