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Proof that SU(n) is a group

  1. Nov 6, 2015 #1
    1. The problem statement, all variables and given/known data

    Show that the set of all ##n \times n## unitary matrices with unit determinant forms a group.

    2. Relevant equations


    3. The attempt at a solution

    For two unitary matrices ##U_{1}## and ##U_{2}## with unit determinant, det(##U_{1}U_{2}##) = det(##U_{1}##)det(##U_{2}##) = 1.

    So, closure is obeyed.

    Matrix multiplication is associative.

    The identity matrix is unitary with unit determinant.

    For a unitary matrix ##U_{1}## with unit determinant, 1 = det(##U_{1}U_{1}^{-1}##) = det(##(U_{1}##)det(##U_{1})^{-1}##) = det(##U_{1})^{-1}##).

    Therefore, the inverse exists.

    Am I correct?
     
  2. jcsd
  3. Nov 6, 2015 #2

    fresh_42

    Staff: Mentor

    Your first version was better. You just had to add that 1 is unitary and unitary matrices are invertible.
     
  4. Nov 6, 2015 #3
    How does the fact that 1 is unitary prove anything? :frown:

    Same goes for the fact that the unitary matrices are invertible.
     
  5. Nov 6, 2015 #4

    fresh_42

    Staff: Mentor

    You need to have 1 inside the group. And you showed that the inverse is unitary. What you did not show was, that the inverse exists at all. Of course it does, since det U cannot be zero. (As I said, it was almost perfect already. These 2 facts I mentioned are a little sophisticated I admit. But in a strong sense it has to be mentioned.)
     
  6. Nov 6, 2015 #5
    Do I need to have 1 in the group because it is the identity element? Is that what you mean?
     
  7. Nov 6, 2015 #6

    fresh_42

    Staff: Mentor

    Yes. I meant the unity matrix with 1. It all depends on your definition of SU(n). If you require SU(n) ⊂ GL(n) to be "the set of all invertible (n x n) matrices that ..." then you're done with the inverse element by showing it is unitary. If you require SU(n) ⊂ M(n) to be "the set of any (n x n) matrices that ..." then it's not clear a priori that unitary matrices are invertible. Perhaps U^-1 doesn't exits. You perfectly have done the hard part and showed that if it exists, then it has to be unitary, i.e. within the group. The existence hasn't been mentioned. I mean there are matrices that cannot be inverted.
    The existence of the unitary matrix, the 1 in GL(n) is clear. Of course it is also unitary so 1 is even in SU(n). Just mention it. Otherwise you would have a 1 but not as part of the group. It has to be - an clearly is - in the group.
     
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