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Danos
Danos
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As it was pointed out in the question, for $G$ abelian, the simple objects in $\mathcal{Z}(\mathrm{Vect}_G)$ are parametrized by the group $G\oplus G^\vee$ where $G$ is the dual character group $$G^\vee:= \mathrm{Hom}(G,\mathbb{k}^\times).$$

(In general, one would have pairs $(c,V)$ of conjugacy classes $c$ and irreps of the corresponding centraliser.)

To understand what the $S$-matrix does, we need to understand the braiding. Given two simple objecst $(g,\chi)$ and $(h,\tilde{\chi})\in G\oplus G^\vee$ check that the braiding (determined by half-braidings) is given by $$ c_{(g,\chi), (h,\tilde{\chi})} = \chi(h).$$ In particular, their double braiding is: \begin{equation} c_{(h,\tilde{\chi}),(g,\chi)}c_{(g,\chi), (h,\tilde{\chi})} = \chi(h)\tilde{\chi}(g). \end{equation}

Therefore, an object $(g,\chi)$ is transparent if and only if $\chi(h)\tilde{\chi}(g) = 1$ for all $(h,\tilde{h})\in G\oplus G^\vee$$(h,\tilde{\chi})\in G\oplus G^\vee$, which only holds for $(g,\chi) = (1,1)$. Hence $\mathcal{Z}(\mathrm{Vect}_G)$ is non-degenerate.

Let me also give a reference EGNO and point to Section 8.4 (and Example 8.5.4). The category $\mathcal{Z}(\mathrm{Vect}_G)$ is (braided) equivalent to the category $\mathcal{C}(G \oplus G^\vee,q)$ of the metric group $(G\oplus G^\vee,q)$ where $q:G\oplus G^\vee \rightarrow \mathbb{k}^\times$ is the quadratic form $q(g,\chi):= \chi(g)$.

(Through S-matrix)

The $S$-matrix read from above is $$S_{(g,\chi),(h,\tilde{\chi})} = \chi(h)\tilde{\chi}(g).$$ Note that you can make it even more explicit if you fix an iso $G\cong \mathbb{Z}_{n_1}\times\dots \mathbb{Z}_{n_k}$ (Exercise).

Showing that $S$ is invertible is equivalent to showing that $S^2$ is invertible. The following follows from standard computations (orthogonality) in character theory.

\begin{align} S^2_{(g,\chi),(h,\tilde{\chi})} &= \sum_{(k,\chi')}{\chi(k)\tilde{\chi}(k)\chi'(gh)}\\ &= |G| \delta_{g,h^{-1}}\delta_{\chi,\tilde{\chi}^-1}~ \end{align} where the last equality uses that $\chi^{-1} = \overline{\chi}$ and both orthogonality of characters and orthogonality of columns. The result is the conjugation matrix (up to $|G|$). In particular $S^2$ is invertible as $S^4= |G|^2 1$.

As it was pointed out in the question, for $G$ abelian, the simple objects in $\mathcal{Z}(\mathrm{Vect}_G)$ are parametrized by the group $G\oplus G^\vee$ where $G$ is the dual character group $$G^\vee:= \mathrm{Hom}(G,\mathbb{k}^\times).$$

(In general, one would have pairs $(c,V)$ of conjugacy classes $c$ and irreps of the corresponding centraliser.)

To understand what the $S$-matrix does, we need to understand the braiding. Given two simple objecst $(g,\chi)$ and $(h,\tilde{\chi})\in G\oplus G^\vee$ check that the braiding (determined by half-braidings) is given by $$ c_{(g,\chi), (h,\tilde{\chi})} = \chi(h).$$ In particular, their double braiding is: \begin{equation} c_{(h,\tilde{\chi}),(g,\chi)}c_{(g,\chi), (h,\tilde{\chi})} = \chi(h)\tilde{\chi}(g). \end{equation}

Therefore, an object $(g,\chi)$ is transparent if and only if $\chi(h)\tilde{\chi}(g) = 1$ for all $(h,\tilde{h})\in G\oplus G^\vee$, which only holds for $(g,\chi) = (1,1)$. Hence $\mathcal{Z}(\mathrm{Vect}_G)$ is non-degenerate.

Let me also give a reference EGNO and point to Section 8.4 (and Example 8.5.4). The category $\mathcal{Z}(\mathrm{Vect}_G)$ is (braided) equivalent to the category $\mathcal{C}(G \oplus G^\vee,q)$ of the metric group $(G\oplus G^\vee,q)$ where $q:G\oplus G^\vee \rightarrow \mathbb{k}^\times$ is the quadratic form $q(g,\chi):= \chi(g)$.

(Through S-matrix)

The $S$-matrix read from above is $$S_{(g,\chi),(h,\tilde{\chi})} = \chi(h)\tilde{\chi}(g).$$ Note that you can make it even more explicit if you fix an iso $G\cong \mathbb{Z}_{n_1}\times\dots \mathbb{Z}_{n_k}$ (Exercise).

Showing that $S$ is invertible is equivalent to showing that $S^2$ is invertible. The following follows from standard computations (orthogonality) in character theory.

\begin{align} S^2_{(g,\chi),(h,\tilde{\chi})} &= \sum_{(k,\chi')}{\chi(k)\tilde{\chi}(k)\chi'(gh)}\\ &= |G| \delta_{g,h^{-1}}\delta_{\chi,\tilde{\chi}^-1}~ \end{align} where the last equality uses that $\chi^{-1} = \overline{\chi}$ and both orthogonality of characters and orthogonality of columns. The result is the conjugation matrix (up to $|G|$). In particular $S^2$ is invertible as $S^4= |G|^2 1$.

As it was pointed out in the question, for $G$ abelian, the simple objects in $\mathcal{Z}(\mathrm{Vect}_G)$ are parametrized by the group $G\oplus G^\vee$ where $G$ is the dual character group $$G^\vee:= \mathrm{Hom}(G,\mathbb{k}^\times).$$

(In general, one would have pairs $(c,V)$ of conjugacy classes $c$ and irreps of the corresponding centraliser.)

To understand what the $S$-matrix does, we need to understand the braiding. Given two simple objecst $(g,\chi)$ and $(h,\tilde{\chi})\in G\oplus G^\vee$ check that the braiding (determined by half-braidings) is given by $$ c_{(g,\chi), (h,\tilde{\chi})} = \chi(h).$$ In particular, their double braiding is: \begin{equation} c_{(h,\tilde{\chi}),(g,\chi)}c_{(g,\chi), (h,\tilde{\chi})} = \chi(h)\tilde{\chi}(g). \end{equation}

Therefore, an object $(g,\chi)$ is transparent if and only if $\chi(h)\tilde{\chi}(g) = 1$ for all $(h,\tilde{\chi})\in G\oplus G^\vee$, which only holds for $(g,\chi) = (1,1)$. Hence $\mathcal{Z}(\mathrm{Vect}_G)$ is non-degenerate.

Let me also give a reference EGNO and point to Section 8.4 (and Example 8.5.4). The category $\mathcal{Z}(\mathrm{Vect}_G)$ is (braided) equivalent to the category $\mathcal{C}(G \oplus G^\vee,q)$ of the metric group $(G\oplus G^\vee,q)$ where $q:G\oplus G^\vee \rightarrow \mathbb{k}^\times$ is the quadratic form $q(g,\chi):= \chi(g)$.

(Through S-matrix)

The $S$-matrix read from above is $$S_{(g,\chi),(h,\tilde{\chi})} = \chi(h)\tilde{\chi}(g).$$ Note that you can make it even more explicit if you fix an iso $G\cong \mathbb{Z}_{n_1}\times\dots \mathbb{Z}_{n_k}$ (Exercise).

Showing that $S$ is invertible is equivalent to showing that $S^2$ is invertible. The following follows from standard computations (orthogonality) in character theory.

\begin{align} S^2_{(g,\chi),(h,\tilde{\chi})} &= \sum_{(k,\chi')}{\chi(k)\tilde{\chi}(k)\chi'(gh)}\\ &= |G| \delta_{g,h^{-1}}\delta_{\chi,\tilde{\chi}^-1}~ \end{align} where the last equality uses that $\chi^{-1} = \overline{\chi}$ and both orthogonality of characters and orthogonality of columns. The result is the conjugation matrix (up to $|G|$). In particular $S^2$ is invertible as $S^4= |G|^2 1$.

Added the S-matrix segment in request to Xiaomeng's comment.
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Danos
Danos
  • 166
  • 4

As it was pointed out in the question, for $G$ abelian, the simple objects in $\mathcal{Z}(\mathrm{Vect}_G)$ are parametrized by the group $G\oplus G^\vee$ where $G$ is the dual character group $$G^\vee:= \mathrm{Hom}(G,\mathbb{k}^\times).$$

(In general, one would have pairs $(c,V)$ of conjugacy classes $c$ and irreps of the corresponding centraliser.)

To understand what the $S$-matrix does, we need to understand the braiding. Given two simple objecst $(g,\chi)$ and $(h,\tilde{\chi})\in G\oplus G^\vee$ check that the braiding (determined by half-braidings) is given by $$ c_{(g,\chi), (h,\tilde{\chi})} = \chi(h).$$ In particular, their double braiding is: \begin{equation} c_{(h,\tilde{\chi}),(g,\chi)}c_{(g,\chi), (h,\tilde{\chi})} = \chi(h)\tilde{\chi}(g). \end{equation}

Therefore, an object $(g,\chi)$ is transparent if and only if $\chi(h)\tilde{\chi}(g) = 1$ for all $(h,\tilde{h})\in G\oplus G^\vee$, which only holds for $(g,\chi) = (1,1)$. Hence $\mathcal{Z}(\mathrm{Vect}_G)$ is non-degenerate.

Let me also give a reference EGNO and point to Section 8.4 (and Example 8.5.4). The category $\mathcal{Z}(\mathrm{Vect}_G)$ is (braided) equivalent to the category $\mathcal{C}(G \oplus G^\vee,q)$ of the metric group $(G\oplus G^\vee,q)$ where $q:G\oplus G^\vee \rightarrow \mathbb{k}^\times$ is the quadratic form $q(g,\chi):= \chi(g)$.

(Through S-matrix)

The $S$-matrix read from above is $$S_{(g,\chi),(h,\tilde{\chi})} = \chi(h)\tilde{\chi}(g).$$ Note that you can make it even more explicit if you fix an iso $G\cong \mathbb{Z}_{n_1}\times\dots \mathbb{Z}_{n_k}$ (Exercise).

Showing that $S$ is invertible is equivalent to showing that $S^2$ is invertible. The following follows from standard computations (orthogonality) in character theory.

\begin{align} S^2_{(g,\chi),(h,\tilde)} &= \sum_{(k,\chi')}{\chi(k)\tilde{\chi}(k)\chi'(gh)}\\ &= |G| \delta_{g,h^{-1}}\delta_{\chi,\tilde{\chi}^-1}~ \end{align}\begin{align} S^2_{(g,\chi),(h,\tilde{\chi})} &= \sum_{(k,\chi')}{\chi(k)\tilde{\chi}(k)\chi'(gh)}\\ &= |G| \delta_{g,h^{-1}}\delta_{\chi,\tilde{\chi}^-1}~ \end{align} where the last equality uses that $\chi^{-1} = \overline{\chi}$ and both orthogonality of characters and orthogonality of columns. The result is the \textit{conjugation matrix}conjugation matrix (up to $|G|$). In particular $S^2$ is invertible as $S^4= |G|^2 1$.

As it was pointed out in the question, for $G$ abelian, the simple objects in $\mathcal{Z}(\mathrm{Vect}_G)$ are parametrized by the group $G\oplus G^\vee$ where $G$ is the dual character group $$G^\vee:= \mathrm{Hom}(G,\mathbb{k}^\times).$$

(In general, one would have pairs $(c,V)$ of conjugacy classes $c$ and irreps of the corresponding centraliser.)

To understand what the $S$-matrix does, we need to understand the braiding. Given two simple objecst $(g,\chi)$ and $(h,\tilde{\chi})\in G\oplus G^\vee$ check that the braiding (determined by half-braidings) is given by $$ c_{(g,\chi), (h,\tilde{\chi})} = \chi(h).$$ In particular, their double braiding is: \begin{equation} c_{(h,\tilde{\chi}),(g,\chi)}c_{(g,\chi), (h,\tilde{\chi})} = \chi(h)\tilde{\chi}(g). \end{equation}

Therefore, an object $(g,\chi)$ is transparent if and only if $\chi(h)\tilde{\chi}(g) = 1$ for all $(h,\tilde{h})\in G\oplus G^\vee$, which only holds for $(g,\chi) = (1,1)$. Hence $\mathcal{Z}(\mathrm{Vect}_G)$ is non-degenerate.

Let me also give a reference EGNO and point to Section 8.4 (and Example 8.5.4). The category $\mathcal{Z}(\mathrm{Vect}_G)$ is (braided) equivalent to the category $\mathcal{C}(G \oplus G^\vee,q)$ of the metric group $(G\oplus G^\vee,q)$ where $q:G\oplus G^\vee \rightarrow \mathbb{k}^\times$ is the quadratic form $q(g,\chi):= \chi(g)$.

(Through S-matrix)

The $S$-matrix read from above is $$S_{(g,\chi),(h,\tilde{\chi})} = \chi(h)\tilde{\chi}(g).$$ Note that you can make it even more explicit if you fix an iso $G\cong \mathbb{Z}_{n_1}\times\dots \mathbb{Z}_{n_k}$ (Exercise).

Showing that $S$ is invertible is equivalent to showing that $S^2$ is invertible. The following follows from standard computations (orthogonality) in character theory.

\begin{align} S^2_{(g,\chi),(h,\tilde)} &= \sum_{(k,\chi')}{\chi(k)\tilde{\chi}(k)\chi'(gh)}\\ &= |G| \delta_{g,h^{-1}}\delta_{\chi,\tilde{\chi}^-1}~ \end{align} where the last equality uses that $\chi^{-1} = \overline{\chi}$ and both orthogonality of characters and orthogonality of columns. The result is the \textit{conjugation matrix} (up to $|G|$). In particular $S^2$ is invertible as $S^4= |G|^2 1$.

As it was pointed out in the question, for $G$ abelian, the simple objects in $\mathcal{Z}(\mathrm{Vect}_G)$ are parametrized by the group $G\oplus G^\vee$ where $G$ is the dual character group $$G^\vee:= \mathrm{Hom}(G,\mathbb{k}^\times).$$

(In general, one would have pairs $(c,V)$ of conjugacy classes $c$ and irreps of the corresponding centraliser.)

To understand what the $S$-matrix does, we need to understand the braiding. Given two simple objecst $(g,\chi)$ and $(h,\tilde{\chi})\in G\oplus G^\vee$ check that the braiding (determined by half-braidings) is given by $$ c_{(g,\chi), (h,\tilde{\chi})} = \chi(h).$$ In particular, their double braiding is: \begin{equation} c_{(h,\tilde{\chi}),(g,\chi)}c_{(g,\chi), (h,\tilde{\chi})} = \chi(h)\tilde{\chi}(g). \end{equation}

Therefore, an object $(g,\chi)$ is transparent if and only if $\chi(h)\tilde{\chi}(g) = 1$ for all $(h,\tilde{h})\in G\oplus G^\vee$, which only holds for $(g,\chi) = (1,1)$. Hence $\mathcal{Z}(\mathrm{Vect}_G)$ is non-degenerate.

Let me also give a reference EGNO and point to Section 8.4 (and Example 8.5.4). The category $\mathcal{Z}(\mathrm{Vect}_G)$ is (braided) equivalent to the category $\mathcal{C}(G \oplus G^\vee,q)$ of the metric group $(G\oplus G^\vee,q)$ where $q:G\oplus G^\vee \rightarrow \mathbb{k}^\times$ is the quadratic form $q(g,\chi):= \chi(g)$.

(Through S-matrix)

The $S$-matrix read from above is $$S_{(g,\chi),(h,\tilde{\chi})} = \chi(h)\tilde{\chi}(g).$$ Note that you can make it even more explicit if you fix an iso $G\cong \mathbb{Z}_{n_1}\times\dots \mathbb{Z}_{n_k}$ (Exercise).

Showing that $S$ is invertible is equivalent to showing that $S^2$ is invertible. The following follows from standard computations (orthogonality) in character theory.

\begin{align} S^2_{(g,\chi),(h,\tilde{\chi})} &= \sum_{(k,\chi')}{\chi(k)\tilde{\chi}(k)\chi'(gh)}\\ &= |G| \delta_{g,h^{-1}}\delta_{\chi,\tilde{\chi}^-1}~ \end{align} where the last equality uses that $\chi^{-1} = \overline{\chi}$ and both orthogonality of characters and orthogonality of columns. The result is the conjugation matrix (up to $|G|$). In particular $S^2$ is invertible as $S^4= |G|^2 1$.

Added the S-matrix segment in request to Xiaomeng's comment.
Source Link
Danos
Danos
  • 166
  • 4

As it was pointed out in the question, for $G$ abelian, the simple objects in $\mathcal{Z}(\mathrm{Vect}_G)$ are parametrized by the group $G\oplus G^\vee$ where $G$ is the dual character group $$G^\vee:= \mathrm{Hom}(G,\mathbb{k}^\times).$$

(In general, one would have pairs $(c,V)$ of conjugacy classes $c$ and irreps of the corresponding centraliser.)

To understand what the $S$-matrix does, we need to understand the braiding. Given two simple objecst $(g,\chi)$ and $(h,\tilde{\chi})\in G\oplus G^\vee$ check that the braiding (determined by half-braidings) is given by $$ c_{(g,\chi), (h,\tilde{\chi})} = \chi(h).$$ In particular, their double braiding is: \begin{equation} c_{(h,\tilde{\chi}),(g,\chi)}c_{(g,\chi), (h,\tilde{\chi})} = \chi(h)\tilde{\chi}(g). \end{equation}

Therefore, an object $(g,\chi)$ is transparent if and only if $\chi(h)\tilde{\chi}(g) = 1$ for all $(h,\tilde{h})\in G\oplus G^\vee$, which only holds for $(g,\chi) = (1,1)$. Hence $\mathcal{Z}(\mathrm{Vect}_G)$ is non-degenerate.

Let me also give a reference EGNO and point to Section 8.4 (and Example 8.5.4). The category $\mathcal{Z}(\mathrm{Vect}_G)$ is (braided) equivalent to the category $\mathcal{C}(G \oplus G^\vee,q)$ of the metric group $(G\oplus G^\vee,q)$ where $q:G\oplus G^\vee \rightarrow \mathbb{k}^\times$ is the quadratic form $q(g,\chi):= \chi(g)$.

(Through S-matrix)

The $S$-matrix read from above is $$S_{(g,\chi),(h,\tilde{\chi})} = \chi(h)\tilde{\chi}(g).$$ Note that you can make it even more explicit if you fix an iso $G\cong \mathbb{Z}_{n_1}\times\dots \mathbb{Z}_{n_k}$ (Exercise).

Showing that $S$ is invertible is equivalent to showing that $S^2$ is invertible. The following follows from standard computations (orthogonality) in character theory.

\begin{align} S^2_{(g,\chi),(h,\tilde)} &= \sum_{(k,\chi')}{\chi(k)\tilde{\chi}(k)\chi'(gh)}\\ &= |G| \delta_{g,h^{-1}}\delta_{\chi,\tilde{\chi}^-1}~ \end{align} where the last equality uses that $\chi^{-1} = \overline{\chi}$ and both orthogonality of characters and orthogonality of columns. The result is the \textit{conjugation matrix} (up to $|G|$). In particular $S^2$ is invertible as $S^4= |G|^2 1$.

As it was pointed out in the question, for $G$ abelian, the simple objects in $\mathcal{Z}(\mathrm{Vect}_G)$ are parametrized by the group $G\oplus G^\vee$ where $G$ is the dual character group $$G^\vee:= \mathrm{Hom}(G,\mathbb{k}^\times).$$

(In general, one would have pairs $(c,V)$ of conjugacy classes $c$ and irreps of the corresponding centraliser.)

To understand what the $S$-matrix does, we need to understand the braiding. Given two simple objecst $(g,\chi)$ and $(h,\tilde{\chi})\in G\oplus G^\vee$ check that the braiding (determined by half-braidings) is given by $$ c_{(g,\chi), (h,\tilde{\chi})} = \chi(h).$$ In particular, their double braiding is: \begin{equation} c_{(h,\tilde{\chi}),(g,\chi)}c_{(g,\chi), (h,\tilde{\chi})} = \chi(h)\tilde{\chi}(g). \end{equation}

Therefore, an object $(g,\chi)$ is transparent if and only if $\chi(h)\tilde{\chi}(g) = 1$ for all $(h,\tilde{h})\in G\oplus G^\vee$, which only holds for $(g,\chi) = (1,1)$. Hence $\mathcal{Z}(\mathrm{Vect}_G)$ is non-degenerate.

Let me also give a reference EGNO and point to Section 8.4 (and Example 8.5.4). The category $\mathcal{Z}(\mathrm{Vect}_G)$ is (braided) equivalent to the category $\mathcal{C}(G \oplus G^\vee,q)$ of the metric group $(G\oplus G^\vee,q)$ where $q:G\oplus G^\vee \rightarrow \mathbb{k}^\times$ is the quadratic form $q(g,\chi):= \chi(g)$.

As it was pointed out in the question, for $G$ abelian, the simple objects in $\mathcal{Z}(\mathrm{Vect}_G)$ are parametrized by the group $G\oplus G^\vee$ where $G$ is the dual character group $$G^\vee:= \mathrm{Hom}(G,\mathbb{k}^\times).$$

(In general, one would have pairs $(c,V)$ of conjugacy classes $c$ and irreps of the corresponding centraliser.)

To understand what the $S$-matrix does, we need to understand the braiding. Given two simple objecst $(g,\chi)$ and $(h,\tilde{\chi})\in G\oplus G^\vee$ check that the braiding (determined by half-braidings) is given by $$ c_{(g,\chi), (h,\tilde{\chi})} = \chi(h).$$ In particular, their double braiding is: \begin{equation} c_{(h,\tilde{\chi}),(g,\chi)}c_{(g,\chi), (h,\tilde{\chi})} = \chi(h)\tilde{\chi}(g). \end{equation}

Therefore, an object $(g,\chi)$ is transparent if and only if $\chi(h)\tilde{\chi}(g) = 1$ for all $(h,\tilde{h})\in G\oplus G^\vee$, which only holds for $(g,\chi) = (1,1)$. Hence $\mathcal{Z}(\mathrm{Vect}_G)$ is non-degenerate.

Let me also give a reference EGNO and point to Section 8.4 (and Example 8.5.4). The category $\mathcal{Z}(\mathrm{Vect}_G)$ is (braided) equivalent to the category $\mathcal{C}(G \oplus G^\vee,q)$ of the metric group $(G\oplus G^\vee,q)$ where $q:G\oplus G^\vee \rightarrow \mathbb{k}^\times$ is the quadratic form $q(g,\chi):= \chi(g)$.

(Through S-matrix)

The $S$-matrix read from above is $$S_{(g,\chi),(h,\tilde{\chi})} = \chi(h)\tilde{\chi}(g).$$ Note that you can make it even more explicit if you fix an iso $G\cong \mathbb{Z}_{n_1}\times\dots \mathbb{Z}_{n_k}$ (Exercise).

Showing that $S$ is invertible is equivalent to showing that $S^2$ is invertible. The following follows from standard computations (orthogonality) in character theory.

\begin{align} S^2_{(g,\chi),(h,\tilde)} &= \sum_{(k,\chi')}{\chi(k)\tilde{\chi}(k)\chi'(gh)}\\ &= |G| \delta_{g,h^{-1}}\delta_{\chi,\tilde{\chi}^-1}~ \end{align} where the last equality uses that $\chi^{-1} = \overline{\chi}$ and both orthogonality of characters and orthogonality of columns. The result is the \textit{conjugation matrix} (up to $|G|$). In particular $S^2$ is invertible as $S^4= |G|^2 1$.

deleted 7 characters in body
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Danos
Danos
  • 166
  • 4
Source Link
Danos
Danos
  • 166
  • 4