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    • CommentRowNumber1.
    • CommentAuthorJohn Baez
    • CommentTimeJun 26th 2010

    I added material to Young diagram, which forced me to create entries for special linear group and special unitary group. I also added a slight clarification to unitary group.

    I would love it if someone who knows algebraic geometry would fix this remark at general linear group:

    Given a commutative field kk, the general linear group GL(n,k)GL(n,k) (or GL n(k)GL_n(k)) is the group of invertible n×nn\times n matrices with entries in kk. It can be considered as a subvariety of the affine space M n×n(k)M_{n\times n}(k) of square matrices of size nn carved out by the equations saying that the determinant of a matrix is zero.

    In fact it’s ’carved out’ by the inequality saying the determinant is not zero… so its description as an algebraic variety is somewhat different than suggested above. Right???

    • CommentRowNumber2.
    • CommentAuthorTodd_Trimble
    • CommentTimeJun 26th 2010

    Yes, as a variety you could define it by the equation det(M)t=1det(M)t = 1 where tt is an extra indeterminate you throw in.

    • CommentRowNumber3.
    • CommentAuthorHarry Gindi
    • CommentTimeJun 26th 2010
    • (edited Jun 26th 2010)

    Alright, the actual group scheme is GL nGL_n, which is the functor of points sending each commutative ring to the group of invertible n×nn\times n matrices over that ring.

    GL n(R)GL_n(R) is secretly notation for GL n× Spec(Z)Spec(R)GL_n\times_{Spec(\mathbf{Z})}Spec(R), which is the fibre of GL nGL_n over RR.

    Actually proving it’s a scheme requires actually knowing something about algebraic geometry (although it probably isn’t very hard. I think Toen gives a proof of the special case for GL 1GL_1), however, so I’ll take my leave.

    • CommentRowNumber4.
    • CommentAuthorTodd_Trimble
    • CommentTimeJun 26th 2010

    Isn’t it just an affine scheme given by the commutative ring [x 11,,x nn,t]/(det(X)t1)\mathbb{Z}[x_{11}, \ldots, x_{nn}, t]/(det(X)t - 1) ? Am I making a stupid mistake?

    • CommentRowNumber5.
    • CommentAuthorHarry Gindi
    • CommentTimeJun 26th 2010
    • (edited Jun 26th 2010)

    I think so. I guess it was easier than I thought!

    • CommentRowNumber6.
    • CommentAuthorJohn Baez
    • CommentTimeJun 26th 2010

    I’m sure you’re right, Todd - I just want someone to step in and lay claim to the title of ’someone who knows algebraic geometry’ by correcting that darn mistake. I’m too busy doing other stuff.

    (Read my queries on the Schur functor page!)

    • CommentRowNumber7.
    • CommentAuthorTodd_Trimble
    • CommentTimeJun 26th 2010

    John, I fixed the glitch at general linear group, and I’ve read through the queries at Schur functor, which I’ll answer when I get a suitable moment. I’m having fun with that!

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