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It seems intuitively obvious to me that there cannot be an isomorphism between $\mathrm{SU}(2)$ and $\mathrm{SU}(2)\times\mathrm{SU}(2)$ where SU(2) is the Lie Group with the Pauli matrices as generators and $\mathrm{SU}(2)\times\mathrm{SU}(2)$ is the Direct Product whose generators are formed from the generators of $\mathrm{SU}(2)$ by putting them into $2\times2$ matrices in block diagonal form.

(I was told this is a correct method of taking the Direct Product in an answer to a prior question. See the response "that is one method of taking the Direct Product":

http://physics.stackexchange.com/questions/28505/how-do-i-construct-the-su2-representation-of-the-lorentz-group-using-su2)

And, if so, is the following argument valid?

$\mathrm{SU}(2)$ is not isomorphic to $\mathrm{SO}(1,3)$ because $\mathrm{SO}(1,3)$ is isomorphic to $\mathrm{SU}(2)\times \mathrm{SU}(2)$. If $\mathrm{SU}(2)$ is isomorphic to $\mathrm{SO}(1,3)$, then $\mathrm{SU}(2)$ is isomorphic to $\mathrm{SU}(2)\times \mathrm{SU}(2)$. But, that is false, therefore $\mathrm{SU}(2)$ cannot be isomorphic to $\mathrm{SO}(1,3)$.

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2 Answers 2

$SO(3,1)$ is not isomorphic to $SU(2)\times SU(2)$ (notice e.g. that $SU(2)\times SU(2)$ os compact while $SO(3,1)$ is not), but rather to $SL(2,\mathbb{C})/\pm1$. (however $SL(2,\mathbb{C})$ and $SU(2)\times SU(2)$ are two real version of the same complex group $SL(2,\mathbb{C})\times SL(2,\mathbb{C})$.) Anyway, $SO(3,1)$ is not isomorphic to $SU(2)$ simply by dimension count: the first has dimension $6$ while the second has $3$.

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But isn't it the case that any Lorentz Transformation can be found by exponentiating the generators of SU(2)XSU(2)? If any transformation can be described by a group action in SU(2)XSU(2), then how can there not be an isomorphism? How would you characterize the relationship between SU(2)XSU(2) and the Lorentz Group? –  bb6 May 20 '12 at 13:03
    
And could SU(2) be isomorphic to SO(2,1) or O(2,1)? and are these representations of the Lorentz Group of Transformations with only one space component? –  bb6 May 20 '12 at 13:11
    
Also, $SU(2)$ is connected, whereas $SO(3,1)$ isn't. –  M Turgeon May 20 '12 at 15:06
    
@barry I'm not an expert in these things, but here is what I found on Wikipedia. Indeed, you can get generators of $SO(3,1)$ by exponentiating (suitable multiples of) the Pauli matrices. But that does not mean you get isomorphic Lie groups. Another example is the Lie algebras $\mathfrak{su}_2$ and $\mathfrak{so}_3$; they are isomorphic as Lie algebras, but $SU(2)$ is not isomorphic to $SO(3)$ (this has to do with the fact that $SU(2)$ is a double cover of $SO(3)$). –  M Turgeon May 20 '12 at 15:16
    
Thanks for these replies. They were very helpful. I thought groups with the same algebra must be isomorphic to each other. It seemed obvious to me at the time. I'll be more careful to distinguish Lie Groups from their Algebras in the future. I've been working from a physics textbook that does not provide much information about group theory. Could anyone recommend a good group theory book? Or a good source on the internet? –  bb6 May 20 '12 at 22:56

barry, perhaps some more comments about the differences between mathematicians and physicists are in order. In my experience, physicists frequently do not distinguish between Lie groups and their Lie algebras, whereas mathematicians are generally careful to make this distinction. There are many pairs of Lie groups which are not isomorphic but which have isomorphic Lie algebras, and for many purposes this is "close enough."

How would you characterize the relationship between SU(2)XSU(2) and the Lorentz Group?

They have isomorphic Lie algebras.

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