The dimension of $operatornameSL(n,F)$ as a linear algebraic group

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For an algebraically closed field $F$, what is the dimension of $operatornameSL(n,F)$ as an algebraic group? Can anyone refer me to a place in the literature where this is calculated?







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    For an algebraically closed field $F$, what is the dimension of $operatornameSL(n,F)$ as an algebraic group? Can anyone refer me to a place in the literature where this is calculated?







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      For an algebraically closed field $F$, what is the dimension of $operatornameSL(n,F)$ as an algebraic group? Can anyone refer me to a place in the literature where this is calculated?







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      For an algebraically closed field $F$, what is the dimension of $operatornameSL(n,F)$ as an algebraic group? Can anyone refer me to a place in the literature where this is calculated?









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      edited Aug 26 at 10:42









      Jendrik Stelzner

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      asked Aug 26 at 10:40









      Orpheus

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          The dimension is $n^2-1$. It's a hypersurface in $F^n^2$ defined
          by the equation determinant$=1$. Any text on linear algebraic groups,
          for instance that by Humphreys, will have this.






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          • Thanks for the answer. Do you know what the dimension of $Sp(2n,F)$ is?
            – Orpheus
            Aug 27 at 3:43










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          1 Answer
          1






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          active

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          up vote
          2
          down vote



          accepted










          The dimension is $n^2-1$. It's a hypersurface in $F^n^2$ defined
          by the equation determinant$=1$. Any text on linear algebraic groups,
          for instance that by Humphreys, will have this.






          share|cite|improve this answer




















          • Thanks for the answer. Do you know what the dimension of $Sp(2n,F)$ is?
            – Orpheus
            Aug 27 at 3:43














          up vote
          2
          down vote



          accepted










          The dimension is $n^2-1$. It's a hypersurface in $F^n^2$ defined
          by the equation determinant$=1$. Any text on linear algebraic groups,
          for instance that by Humphreys, will have this.






          share|cite|improve this answer




















          • Thanks for the answer. Do you know what the dimension of $Sp(2n,F)$ is?
            – Orpheus
            Aug 27 at 3:43












          up vote
          2
          down vote



          accepted







          up vote
          2
          down vote



          accepted






          The dimension is $n^2-1$. It's a hypersurface in $F^n^2$ defined
          by the equation determinant$=1$. Any text on linear algebraic groups,
          for instance that by Humphreys, will have this.






          share|cite|improve this answer












          The dimension is $n^2-1$. It's a hypersurface in $F^n^2$ defined
          by the equation determinant$=1$. Any text on linear algebraic groups,
          for instance that by Humphreys, will have this.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Aug 26 at 10:45









          Lord Shark the Unknown

          88.2k955115




          88.2k955115











          • Thanks for the answer. Do you know what the dimension of $Sp(2n,F)$ is?
            – Orpheus
            Aug 27 at 3:43
















          • Thanks for the answer. Do you know what the dimension of $Sp(2n,F)$ is?
            – Orpheus
            Aug 27 at 3:43















          Thanks for the answer. Do you know what the dimension of $Sp(2n,F)$ is?
          – Orpheus
          Aug 27 at 3:43




          Thanks for the answer. Do you know what the dimension of $Sp(2n,F)$ is?
          – Orpheus
          Aug 27 at 3:43

















           

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