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Are abelian nondegeneratenon-degenerate tensor categories semisimple?

A pivotal monoidal category is called non-degenerate if the inner product (x,y) = Tr(xy*)$\left(x,y\right) = Tr\left(xy^{*}\right)$ (where y*$y^{*}$ is the dual map) is non-degenerate. As a rule of thumb non-degenerate is closely related to semisimplicity. For example, if a category is semisimple then it is automatically non-degenerate (this follows from the fact that simple objects don't have dimension 0$0$). Another way of stating non-degenerate is that the category has no "negligible morphisms" where a morphism is called negligible if any way of composing it in order to get an endomorphism of the trivial gives you the zero map.

If you have an abelian pivotal monoidal category which is non-degenerate is it automatically semisimple?

Are abelian nondegenerate tensor categories semisimple?

A pivotal monoidal category is called non-degenerate if the inner product (x,y) = Tr(xy*) (where y* is the dual map) is non-degenerate. As a rule of thumb non-degenerate is closely related to semisimplicity. For example, if a category is semisimple then it is automatically non-degenerate (this follows from the fact that simple objects don't have dimension 0). Another way of stating non-degenerate is that the category has no "negligible morphisms" where a morphism is called negligible if any way of composing it in order to get an endomorphism of the trivial gives you the zero map.

If you have an abelian pivotal monoidal category which is non-degenerate is it automatically semisimple?

Are abelian non-degenerate tensor categories semisimple?

A pivotal monoidal category is called non-degenerate if the inner product $\left(x,y\right) = Tr\left(xy^{*}\right)$ (where $y^{*}$ is the dual map) is non-degenerate. As a rule of thumb non-degenerate is closely related to semisimplicity. For example, if a category is semisimple then it is automatically non-degenerate (this follows from the fact that simple objects don't have dimension $0$). Another way of stating non-degenerate is that the category has no "negligible morphisms" where a morphism is called negligible if any way of composing it in order to get an endomorphism of the trivial gives you the zero map.

If you have an abelian pivotal monoidal category which is non-degenerate is it automatically semisimple?

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Ben Webster
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A pivotal monoidal category is called non-degenerate if the inner product (x,y) = Tr(xy*) (where y* is the dual map) is non-degenerate. As a rule of thumb non-degenerate is closely related to semisimplicity. For example, if a category is semisimple then it is automatically non-degenerate (this follows from the fact that simple objects don't have dimension 0 objects). Another way of stating non-degenerate is that the category has no "negligible morphisms" where a morphism is called negligible if any way of composing it in order to get an endomorphism of the trivial gives you the zero map.

If you have an abelian pivotal monoidal category which is non-degenerate is it automatically semisimple?

A pivotal monoidal category is called non-degenerate if the inner product (x,y) = Tr(xy*) (where y* is the dual map) is non-degenerate. As a rule of thumb non-degenerate is closely related to semisimplicity. For example, if a category is semisimple then it is automatically non-degenerate (this follows from the fact that simple objects don't have dimension 0 objects). Another way of stating non-degenerate is that the category has no "negligible morphisms" where a morphism is called negligible if any way of composing it in order to get an endomorphism of the trivial gives you the zero map.

If you have an abelian pivotal monoidal category which is non-degenerate is it automatically semisimple?

A pivotal monoidal category is called non-degenerate if the inner product (x,y) = Tr(xy*) (where y* is the dual map) is non-degenerate. As a rule of thumb non-degenerate is closely related to semisimplicity. For example, if a category is semisimple then it is automatically non-degenerate (this follows from the fact that simple objects don't have dimension 0). Another way of stating non-degenerate is that the category has no "negligible morphisms" where a morphism is called negligible if any way of composing it in order to get an endomorphism of the trivial gives you the zero map.

If you have an abelian pivotal monoidal category which is non-degenerate is it automatically semisimple?

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Ben Webster
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Noah Snyder
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