Is it possible to prove that R(A) is a rotation in the following way? (Edit: it was kind of a duplicate of the comment from Jos van der Spek, sorry for asking... Just for other viewers I don't remove my comment..) Proof (continued from the video 12:32) Since we know that R(A) is orthogonal, it suffices to show that det(R(A)) is 1. By writing down explicit formula for A M_v A^{-1} = M_{R(A)v}, we have R is a smooth homomorphism from SU(2) to GL(3). Then since SU(2) is connected (since exp map is continuous and su(2) is connected), we can draw a path from I to A in SU(2). Hence, the image of the path under det(R(.)) is connected too. Since det(R(.)) takes only 1 or -1 as image, connectedness shows that det(R(A)) = 1.
So at what point do you actually prove that im R ⊆ SO(3)? Or does that actually readily follow from the fact that o(3) = so(3), and R_* is the induced homomorphism, so det(R(e^A)) = det( exp (R_* (A)) ) and some final step that I'm missing?
It's not that every matrix in SU(2) is of the form e^A for some A ∈ su(2) right? Because then we'd have it: det(R(e^A)) = det( e^(R_*(A)) ) = e^( tr(R_*(A) ) = e⁰ = 1. But that would require exp : su(2) → SU(2) to be surjective, and that doesn't seem to be the case..?
It's something we discussed in class instead of in the video (these videos are just the pre-watching for an actual class, so I had to leave some things to discuss in person!). My preferred proof uses the fact that O(3) has two connected components SU(2) only has one (so the homomorphism can only hit the component of the identity).
@@josvanderspek1403 And you can also prove it this way (indeed su(2)-->SU(2) is surjective, as is exp for any compact group). The fact that there are different proofs makes it a good topic for a class discussion!
@@jonathanevans27 Thanks! This is a _really_ useful video series by the way! So I'm very happy you put in on YT! =D I have found another proof for surjectivity by the way: you can observe that, for σ_i the Pauli matrices, every matrix in SU(2) can be written as [cos θ + iz sin θ (ix + y) sin θ] [(ix - y) sin θ cos θ - iz sin θ] = cos θ Ι + i sin θ (xσ₁ + yσ₂ + zσ₃) = e^(i(xσ₁ + yσ₂ + zσ₃))θ ∈ exp[su(2)].
@Jos van der Spek: No, though there are a number of books which have influenced the way I've presented the material; there's a video at the end of the series which discusses some of these books.
Is it possible to prove that R(A) is a rotation in the following way? (Edit: it was kind of a duplicate of the comment from Jos van der Spek, sorry for asking... Just for other viewers I don't remove my comment..)
Proof (continued from the video 12:32)
Since we know that R(A) is orthogonal, it suffices to show that det(R(A)) is 1.
By writing down explicit formula for A M_v A^{-1} = M_{R(A)v}, we have R is a smooth homomorphism from SU(2) to GL(3).
Then since SU(2) is connected (since exp map is continuous and su(2) is connected), we can draw a path from I to A in SU(2).
Hence, the image of the path under det(R(.)) is connected too. Since det(R(.)) takes only 1 or -1 as image, connectedness shows that det(R(A)) = 1.
Yes, absolutely: this is my favourite way to see it.
Is it possible to post the boardwork to study?
@15:00 I don't really get how you comput the derivative of the left side of this equation, can you elaborate on that?
He just reuses what he computed few line before.
So at what point do you actually prove that im R ⊆ SO(3)? Or does that actually readily follow from the fact that o(3) = so(3), and R_* is the induced homomorphism, so det(R(e^A)) = det( exp (R_* (A)) ) and some final step that I'm missing?
It's not that every matrix in SU(2) is of the form e^A for some A ∈ su(2) right? Because then we'd have it:
det(R(e^A)) = det( e^(R_*(A)) ) = e^( tr(R_*(A) ) = e⁰ = 1.
But that would require exp : su(2) → SU(2) to be surjective, and that doesn't seem to be the case..?
It's something we discussed in class instead of in the video (these videos are just the pre-watching for an actual class, so I had to leave some things to discuss in person!). My preferred proof uses the fact that O(3) has two connected components SU(2) only has one (so the homomorphism can only hit the component of the identity).
@@josvanderspek1403 And you can also prove it this way (indeed su(2)-->SU(2) is surjective, as is exp for any compact group). The fact that there are different proofs makes it a good topic for a class discussion!
@@jonathanevans27 Thanks! This is a _really_ useful video series by the way! So I'm very happy you put in on YT! =D
I have found another proof for surjectivity by the way: you can observe that, for σ_i the Pauli matrices, every matrix in SU(2) can be written as
[cos θ + iz sin θ (ix + y) sin θ]
[(ix - y) sin θ cos θ - iz sin θ] = cos θ Ι + i sin θ (xσ₁ + yσ₂ + zσ₃) = e^(i(xσ₁ + yσ₂ + zσ₃))θ ∈ exp[su(2)].
@@josvanderspek1403 Indeed, that's a nice way to see surjectivity in the case of SU(2)
Are you following a certain book by the way?
@Jos van der Spek: No, though there are a number of books which have influenced the way I've presented the material; there's a video at the end of the series which discusses some of these books.