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I was just curious if anyone knows what kind of constraints one can place on $SL(4,R)$ the set of 4x4 invertible matrices with unit determinant to obtain: $SL(2,C)$ the set of 2x2 invertible complex matrices with unit determinant. I'm guessing that if one places a particular constaint on the former set then they are isomorphic to the latter?

Essentially I'm trying to show that a particular set of matrices I have might be represented in terms of $SL(2,C)$ rather than the current form I have them in $SL(4,R)$. Thanks a ton!

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  • $\begingroup$ This would be easier to answer, I think, if you stated what the current form you have them in is $\endgroup$ – Omnomnomnom Dec 20 '18 at 23:37
  • $\begingroup$ Are you aware of the usual way of representing $GL(n,\Bbb C)$ in $GL(2n,\Bbb R)$? $\endgroup$ – Omnomnomnom Dec 20 '18 at 23:38
  • $\begingroup$ @Omnomnomnom No, I wasn't, but I think the below answer straightened me out! (: $\endgroup$ – R. Rankin Dec 21 '18 at 3:24
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If $\Lambda$ is of the form: $$\Lambda = \begin{bmatrix} a&b&c&d \\ -b&a&-d&c \\ e&f&g&h \\ -f&e&-h&g \end{bmatrix}$$ then we can isomorphically map it to: \begin{bmatrix} a+ib&c+id \\ e+if&g+ih \end{bmatrix}

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  • $\begingroup$ Thank you! That is exactly what I needed!! (: $\endgroup$ – R. Rankin Dec 21 '18 at 3:25
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    $\begingroup$ In terms of Kronecker products (tensor products), we can write this association as $$ \pmatrix{a&c\\e&g} + i\pmatrix{b&d\\f&h} \leftrightarrow \pmatrix{a&c\\e&g}\otimes\pmatrix{1&0\\0&1} + \pmatrix{b&d\\f&h}\otimes \pmatrix{0&1\\-1&0} $$ $\endgroup$ – Omnomnomnom Dec 21 '18 at 5:12
  • $\begingroup$ @Omnomnomnom Thank you, that also actually helps quite a bit. Sorry physicist here trying to learn groups (: $\endgroup$ – R. Rankin Dec 21 '18 at 10:59

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