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1. • CommentRowNumber1.
   • CommentAuthorJohn Baez
   • CommentTimeJun 26th 2010
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   Author: John Baez
   Format: MarkdownItexI 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]] $k$, the **general linear group** $GL(n,k)$ (or $GL_n(k)$) is the group of invertible $n\times n$ matrices with entries in $k$. It can be considered as a sub[[variety]] of the [[affine scheme|affine space]] $M_{n\times n}(k)$ of square matrices of size $n$ 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???

   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 $k$, the general linear group $GL(n,k)$ (or $GL_n(k)$) is the group of invertible $n\times n$ matrices with entries in $k$. It can be considered as a subvariety of the affine space $M_{n\times n}(k)$ of square matrices of size $n$ 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???

2. • CommentRowNumber2.
   • CommentAuthorTodd_Trimble
   • CommentTimeJun 26th 2010
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   Author: Todd_Trimble
   Format: MarkdownItexYes, as a variety you could define it by the equation $det(M)t = 1$ where $t$ is an extra indeterminate you throw in.

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

3. • CommentRowNumber3.
   • CommentAuthorHarry Gindi
   • CommentTimeJun 26th 2010
   • (edited Jun 26th 2010)
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   Author: Harry Gindi
   Format: MarkdownItexAlright, the actual group scheme is $GL_n$, which is the functor of points sending each commutative ring to the _group_ of invertible $n\times n$ matrices over that ring. $GL_n(R)$ is secretly notation for $GL_n\times_{Spec(\mathbf{Z})}Spec(R)$, which is the fibre of $GL_n$ over $R$. 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 $GL_1$), however, so I'll take my leave.

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

   $GL_n(R)$ is secretly notation for $GL_n\times_{Spec(\mathbf{Z})}Spec(R)$, which is the fibre of $GL_n$ over $R$.

   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 $GL_1$), however, so I’ll take my leave.

4. • CommentRowNumber4.
   • CommentAuthorTodd_Trimble
   • CommentTimeJun 26th 2010
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   Author: Todd_Trimble
   Format: MarkdownItexIsn't it just an affine scheme given by the commutative ring $\mathbb{Z}[x_{11}, \ldots, x_{nn}, t]/(det(X)t - 1)$ ? Am I making a stupid mistake?

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

5. • CommentRowNumber5.
   • CommentAuthorHarry Gindi
   • CommentTimeJun 26th 2010
   • (edited Jun 26th 2010)
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   Author: Harry Gindi
   Format: MarkdownItexI think so. I guess it was easier than I thought!

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

6. • CommentRowNumber6.
   • CommentAuthorJohn Baez
   • CommentTimeJun 26th 2010
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   Author: John Baez
   Format: MarkdownItexI'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!)

   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!)

7. • CommentRowNumber7.
   • CommentAuthorTodd_Trimble
   • CommentTimeJun 26th 2010
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   Author: Todd_Trimble
   Format: MarkdownItexJohn, 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!

   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!