Return to Answer

The following is a development along Hari's answer.

As mentionned by Hari, we may assume that $B$ is an irreducible summand of size $m$. It inherits $q$-normality. Let $L$ be the sub-algebra of $M_m(\mathbb C)$ spanned by $B$ and $B^*$.

The subspace $\ker B\cap\ker B^* $ is invariant under $B$ and $B^* $. From irreducibility, we must have $\ker B\cap\ker B^* =\{0\}$. Set $H:=BB^* + B^* B$, which is Hermitian positive definite. From Cayley-Hamilton, $I_m$ is a polynomial in $H$, thus belongs to $L$. Therefore $L$ is unit algebra.

We prove now that $L$ is a simple algebra: let $J\ne(0)$ be a two-sided ideal in $L$. Choose $M\ne0$ in $J$. Then every $M^k$ with $\ge1$ belongs to $J$. By Cayley-Hamilton, we deduce $Tr(M)I_m\in J$. If $J$ is proper, this implies $Tr(M)=0$. Now, $M^* M\in J$ too, because $L$ is invariant under $C\mapsto C^*$ and $J$ is an ideal. Thus $Tr(M^*M)=0$, which is absurd. Therefore $J=L$ and $L$ is simple.

By Wedderburn's theorem, $L$ is isomorphic to $M_r(K)$ for some $r\ge1$, where $K$ is a division ring with $\mathbb C$ in its center. Because $L$ is finitely generated, $K$ must be of finite dimension over $\mathbb C$. From Frobenius' theorem, $K=\mathbb C$. Finally, $L$ is isomorphic to $M_r(\mathbb C)$.

At last, the morphism $M\mapsto M$ from $L$ into $M_m(\mathbb C)$ is a representation. By assumption, it is irreducible. From Corollary 11.6 and Lemma 11.5 of M. Takesaki's Theory of Operator Algebra, I, we obtain $r=m$. Therefore $L=M_m(\mathbb C)$.

By assumption, $\mathcal S_{2q}$ vanishes identically over $L$, thus over $M_m(\mathbb C)$. This implies easily $m\le q$.

The following is a development along Hari's answer.

As mentionned by Hari, we may assume that $B$ is an irreducible summand of size $m$. It inherits $q$-normality. Let $L$ be the sub-algebra of $M_m(\mathbb C)$ spanned by $B$ and $B^*$.

The subspace $\ker B\cap\ker B^* $ is invariant under $B$ and $B^* $. From irreducibility, we must have $\ker B\cap\ker B^* =\{0\}$. Set $H:=BB^* + B^* B$, which is Hermitian positive definite. From Cayley-Hamilton, $I_m$ is a polynomial in $H$, thus belongs to $L$. Therefore $L$ is unit algebra.

We prove now that $L$ is a simple algebra: let $J\ne(0)$ be a two-sided ideal in $L$. Denote $K$ the intersection of all $\ker M\cap\ker M^* $ when $M\ne0$ runs over $J$. Because of finite dimension, it is actually the intersection of finitely many, say for $M_1,\ldots,M_r$. Then $K=\ker \sum_j(M_jM_j^* +M_j^* M_j)$. Because $J$ is an ideal, $K$ is stable under $B$ and $B^*$. Because $B$ is irreducible, one has either $K=(0)$ or $K={\mathbb C}^m$. The latter is impossible, and the former tells us that $J$ contains an invertible element. Therefore $J=L$ and $L$ is simple.

By Wedderburn's theorem, $L$ is isomorphic to $M_r(K)$ for some $r\ge1$, where $K$ is a division ring with $\mathbb C$ in its center. Because $L$ is finitely generated, $K$ must be of finite dimension over $\mathbb C$. From Frobenius' theorem, $K=\mathbb C$. Finally, $L$ is isomorphic to $M_r(\mathbb C)$.

At last, the morphism $M\mapsto M$ from $L$ into $M_m(\mathbb C)$ is a representation. By assumption, it is irreducible. From Corollary 11.6 and Lemma 11.5 of M. Takesaki's Theory of Operator Algebra, I, we obtain $r=m$. Therefore $L=M_m(\mathbb C)$.

By assumption, $\mathcal S_{2q}$ vanishes identically over $L$, thus over $M_m(\mathbb C)$. This implies easily $m\le q$.

The following is a development along Hari's answer.

As mentionned by Hari, we may assume that $B$ is an irreducible summand of size $m$. It inherits $q$-normality. Let $L$ be the sub-algebra of $M_m(\mathbb C)$ spanned by $B$ and $B^*$.

The subspace $\ker B\cap\ker B^* $ is invariant under $B$ and $B^* $. From irreducibility, we must have $\ker B\cap\ker B^* =\{0\}$. Set $H:=BB^* + B^* B$, which is Hermitian positive definite. From Cayley-Hamilton, $I_m$ is a polynomial in $H$, thus belongs to $L$. Therefore $L$ is unit algebra.

We prove now that $L$ is a simple algebra: let $J\ne(0)$ be a two-sided ideal in $L$. Choose $M\ne0$ in $J$. Then every $M^k$ with $\ge1$ belongs to $J$. By Cayley-Hamilton, we deduce $Tr(M)I_m\in J$. If $J$ is proper, this implies $Tr(M)=0$. Now, $M^* M\in J$ too, because $L$ is invariant under $C\mapsto C^*$ and $J$ is an ideal. Thus $Tr(M^*M)=0$, which is absurd. Therefore $J=L$ and $L$ is simple.

By Wedderburn's theorem, $L$ is isomorphic to $M_r(K)$ for some $r\ge1$, where $K$ is a division ring with $\mathbb C$ in its center. Because $L$ is finitely generated, $K$ must be of finite dimension over $\mathbb C$. From Frobenius' theorem, $K=\mathbb C$. Finally, $L$ is isomorphic to $M_r(\mathbb C)$.

At last, the morphism $M\mapsto M$ from $L$ into $M_m(\mathbb C)$ is non-zero. Because both $L$ and $M_m(\mathbb C)$ are simple, they must be isomorphic. In particular, they have the same dimension, therefore $L=M_m(\mathbb C)$.

By assumption, $\mathcal S_{2q}$ vanishes identically over $L$, thus over $M_m(\mathbb C)$. This implies easily $m\le q$.

The following is a development along Hari's answer.

As mentionned by Hari, we may assume that $B$ is an irreducible summand of size $m$. It inherits $q$-normality. Let $L$ be the sub-algebra of $M_m(\mathbb C)$ spanned by $B$ and $B^*$.

The subspace $\ker B\cap\ker B^* $ is invariant under $B$ and $B^* $. From irreducibility, we must have $\ker B\cap\ker B^* =\{0\}$. Set $H:=BB^* + B^* B$, which is Hermitian positive definite. From Cayley-Hamilton, $I_m$ is a polynomial in $H$, thus belongs to $L$. Therefore $L$ is unit algebra.

We prove now that $L$ is a simple algebra: let $J\ne(0)$ be a two-sided ideal in $L$. Choose $M\ne0$ in $J$. Then every $M^k$ with $\ge1$ belongs to $J$. By Cayley-Hamilton, we deduce $Tr(M)I_m\in J$. If $J$ is proper, this implies $Tr(M)=0$. Now, $M^* M\in J$ too, because $L$ is invariant under $C\mapsto C^*$ and $J$ is an ideal. Thus $Tr(M^*M)=0$, which is absurd. Therefore $J=L$ and $L$ is simple.

By Wedderburn's theorem, $L$ is isomorphic to $M_r(K)$ for some $r\ge1$, where $K$ is a division ring with $\mathbb C$ in its center. Because $L$ is finitely generated, $K$ must be of finite dimension over $\mathbb C$. From Frobenius' theorem, $K=\mathbb C$. Finally, $L$ is isomorphic to $M_r(\mathbb C)$.

At last, the morphism $M\mapsto M$ from $L$ into $M_m(\mathbb C)$ is a representation. By assumption, it is irreducible. From Corollary 11.6 and Lemma 11.5 of M. Takesaki's Theory of Operator Algebra, I, we obtain $r=m$. Therefore $L=M_m(\mathbb C)$.

By assumption, $\mathcal S_{2q}$ vanishes identically over $L$, thus over $M_m(\mathbb C)$. This implies easily $m\le q$.

The following is a development along Hari's answer.

As mentionned by Hari, we may assume that $B$ is an irreducible summand of size $m$. It inherits $q$-normality. Let $L$ be the sub-algebra of $M_m(\mathbb C)$ spanned by $B$ and $B^*$.

The subspace $\ker B\cap\ker B^* $ is invariant under $B$ and $B^* $. From irreducibility, we must have $\ker B\cap\ker B^* =\{0\}$. Set $H:=BB^* + B^* B$, which is Hermitian positive definite. From Cayley-Hamilton, $I_m$ is a polynomial in $H$, thus belongs to $L$. Therefore $L$ is unit algebra.

We prove now that $L$ is a simple algebra: let $J\ne(0)$ be a two-sided ideal in $L$. Choose $M\ne0$ in $J$. Then every $M^k$ with $\ge1$ belongs to $J$. By Cayley-Hamilton, we deduce $Tr(M)I_m\in J$. If $J$ is proper, this implies $Tr(M)=0$. Now, $M^* M\in J$ too, because $L$ is invariant under $C\mapsto C^*$ and $J$ is an ideal. Thus $Tr(M^*M)=0$, which is absurd. Therefore $J=L$ and $L$ is simple.

By Wedderburn's theorem, $L$ is isomorphic to $M_r(K)$ for some $r\ge1$, where $K$ is a division ring with $\mathbb C$ in its center. Because $L$ is finitely generated, $K$ must be of finite dimension over $\mathbb C$. From Frobenius' theorem, $K=\mathbb C$. Finally, $L$ is isomorphic to $M_r(\mathbb C)$. From irreducibility, we must have $L=M_m(\mathbb C)$.

By assumption, $\mathcal S_{2q}$ vanishes identically over $L$, thus over $M_m(\mathbb C)$. This implies easily $m\le q$.

The following is a development along Hari's answer.

As mentionned by Hari, we may assume that $B$ is an irreducible summand of size $m$. It inherits $q$-normality. Let $L$ be the sub-algebra of $M_m(\mathbb C)$ spanned by $B$ and $B^*$.

The subspace $\ker B\cap\ker B^* $ is invariant under $B$ and $B^* $. From irreducibility, we must have $\ker B\cap\ker B^* =\{0\}$. Set $H:=BB^* + B^* B$, which is Hermitian positive definite. From Cayley-Hamilton, $I_m$ is a polynomial in $H$, thus belongs to $L$. Therefore $L$ is unit algebra.

We prove now that $L$ is a simple algebra: let $J\ne(0)$ be a two-sided ideal in $L$. Choose $M\ne0$ in $J$. Then every $M^k$ with $\ge1$ belongs to $J$. By Cayley-Hamilton, we deduce $Tr(M)I_m\in J$. If $J$ is proper, this implies $Tr(M)=0$. Now, $M^* M\in J$ too, because $L$ is invariant under $C\mapsto C^*$ and $J$ is an ideal. Thus $Tr(M^*M)=0$, which is absurd. Therefore $J=L$ and $L$ is simple.

By Wedderburn's theorem, $L$ is isomorphic to $M_r(K)$ for some $r\ge1$, where $K$ is a division ring with $\mathbb C$ in its center. Because $L$ is finitely generated, $K$ must be of finite dimension over $\mathbb C$. From Frobenius' theorem, $K=\mathbb C$. Finally, $L$ is isomorphic to $M_r(\mathbb C)$.

At last, the morphism $M\mapsto M$ from $L$ into $M_m(\mathbb C)$ is non-zero. Because both $L$ and $M_m(\mathbb C)$ are simple, they must be isomorphic. In particular, they have the same dimension, therefore $L=M_m(\mathbb C)$.

By assumption, $\mathcal S_{2q}$ vanishes identically over $L$, thus over $M_m(\mathbb C)$. This implies easily $m\le q$.