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In the Artin's book on Algebra, the author stated a theorem (Ch.9, Thm. 2.2):

"A finite subgroup $G$ of $GL(n,\mathbb{C})$ is conjugate to a subgroup of $U(n)$.

Here, $U(n)$ is the unitary group, i.e. if $\langle \,\,, \rangle$ is the standard Hermitian inner product on $\mathbb{C}^n$ given by $\langle (x_1,\cdots,x_n),(y_1,\cdots,y_n)\rangle=\sum_{}{x_i\bar{y_i}}$ then $U(n)=\{A\in GL(n,\mathbb{C}) \colon \langle Av,Aw\rangle = \langle v,w\rangle,\forall v,w\in \mathbb{C}^n \}$.

The proof is: there is a $G$-invariant Hermitian inner product $\langle\,\, , \rangle_1$ on $\mathbb{C}^n$, and consider an orthonormal basis $B_1$ w.r.t this form. If $P$ is the matrix of transformation which changes standard basis to $B_1$, then $PGP^{-1}\leq U(n)$

Question: In the last statement in the proof, the author says that $PGP^{-1}$ is subgroup of unitary group; but, this unitary group is corresponding to the Hermitian inner product $\langle \,\,, \rangle_1$, i.e. it is the group of linear transformations which preserves the inner product $\langle \,\,, \rangle_1$. Why $PGP^{-1}$ should be subgroup of $U(n)$ which is the group of linear transformations which "preserves" the standard Hermitian inner product $\langle \,\,,\rangle$ on $\mathbb{C}^n$?

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To answer the question a slightly different way: choose an orthonormal basis wrt the $G$-invariant Hermitian form $\langle, \rangle_{1}.$ Let this be $\{ u_{i}: 1 \leq i \leq n \}.$ Since the form is $G$-invariant, for any fixed $g \in G,$ we have $\langle gu_{i},gu_{j} \rangle_{1} = \delta_{ij}$ for all $i,j.$ Hence writing the matrix of (the transformation represented by) $g$ with respect to this basis, the columns (and rows) are orthonormal, and the matrix is unitary. So now you have a group of unitary matrices, which is conjugate to the original one via a change of basis matrix.

To answer comment below, which took to long to leave as a comment: It means that if write $gu_{i} = \sum_{j=1}^{n} a_{ij}(g)u_{j},$ where the $a_{ij}(g)$ are complex numbers, then we have for any $r,s$, $$\delta_{rs} = \langle gu_{r},gu_{s} \rangle_{1} = \sum \sum_{p,q= 1}^{n} a_{rp}(g) {\overline a_{sq}(g)} \langle u_{p}, u_{q} \rangle_{1} .$$ But this is $$\sum_{p =1}^{n} a_{rp}(g) { \overline a_{sp}(g)}$$ because $\langle u_{p}, u_{q} \rangle_{1} = \delta_{pq}.$ So what is left is the usual innner product of the rows of the matrix $[a_{ij}(g)].$

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  • $\begingroup$ very simple, and nice.. $\endgroup$ – Groups Jul 30 '12 at 8:39
  • $\begingroup$ in the of $g$ w.r.t. the basic chosen, columns are orthonormal mans "they are orthonormal w.r.t. the inner product $\langle \,,\rangle_1$, isn't it? $\endgroup$ – Groups Jul 30 '12 at 15:26
  • $\begingroup$ No, see the modified answer. The rows of the matrix are mutually orthogonal and of length $1$, so orthonormal wrt usual hermitian form on $\mathbb{C}^{n}.$Likewise for columns. $\endgroup$ – Geoff Robinson Jul 30 '12 at 22:45
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Preserving the $G$-invariant inner product $\langle , \rangle_1$ implies preserving the original $\langle , \rangle .$ To see this requires knowing how $\langle , \rangle_1$ was constructed (via Weyl's averaging trick.) For $u,v \in \mathbb{C}^n,$ $g\in G,$ and $A \in PGP^{-1},$ we have

$\langle A(gu), A(gv) \rangle_1 = \langle gu, gv \rangle_1 =\langle u, v \rangle_1 =\frac{1}{|G|} \displaystyle\sum_{g\in G} \langle u , v \rangle = \langle u, v \rangle .$

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  • $\begingroup$ pithy$\ \ \ \ \ \ \ \ \ $ $\endgroup$ – David Wheeler Jul 30 '12 at 6:47
  • $\begingroup$ The averaging trick surely uses the average $$\frac{1}{|G|} \sum_{g \in G} \langle gu,gv \rangle$$ which is why it ends up $G$-invariant. (Also, it was known long before Weyl in the case of finite groups). $\endgroup$ – Geoff Robinson Jul 30 '12 at 7:41

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