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Qiaochu Yuan
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As in the previous question, I continue to suggest that everything is much cleaner if you think in terms of bimodules. So we'll ask the more general question: in the 2-category $\text{Bim}(k)$ of $k$-algebras, $k$-bimodules, and bimodule homomorphisms, what is an adjoint pair?

The answer is worked out in this blog post, although I'll warn you that in the convention of that blog post composition of 1-morphisms is written in diagrammatic order, which means concretely that the composition of an $(A, B)$-bimodule $M$ and a $(B, C)$-bimodule $N$, in that order, is $M \otimes_B N$. Adopting the usual composition convention switches the meaning of "left" and "right," but this convention makes things work out very nicely as follows.

Proposition: An $(A, B)$-bimodule $M$ has a left adjoint iff it is finitely presented projective as a left $A$-module, in which case its left adjoint is $\text{Hom}_A(M, A)$. It has a right adjoint iff it is finitely presented projective as a right $B$-module, in which case its right adjoint is $\text{Hom}_B(M, B)$.

KV is the special case where we only consider algebras of the form $k^n$ for $k$ a field and $(k^n, k^m)$-bimodules, which concretely can be thought of as $n \times m$ matrices of $k$-vector spaces, where all of the vector spaces are finite-dimensional. These are all finitely presented projective on both sides, and so we conclude that every morphism of KV 2-vector spaces has both a left and a right adjoint, both of which I believe are concretely given by sending a matrix $V_{ij}$ of vector spaces to the transposed and dualized matrix $V_{ji}^{\ast}$, but I haven't checked this in detail.

(Note that because I'm thinking about 1-morphisms as matrices of vector spaces instead of taking a skeleton and identifying vector spaces with their dimensions, when I write down taking the dual I'm making some implicit claims about functoriality and 2-morphisms that I don't need to keep explicit track of. It's actually more work to work with the skeleton because then I need to distinguish a vector space from its dual by talking a lot more explicitly about 2-morphisms, and I sort of have to conjure the 2-morphisms out of thin air rather than letting them naturally arise out of e.g. the dual pairing.)

As in the previous question, I continue to suggest that everything is much cleaner if you think in terms of bimodules. So we'll ask the more general question: in the 2-category $\text{Bim}(k)$ of $k$-algebras, $k$-bimodules, and bimodule homomorphisms, what is an adjoint pair?

The answer is worked out in this blog post, although I'll warn you that in the convention of that blog post composition of 1-morphisms is written in diagrammatic order, which means concretely that the composition of an $(A, B)$-bimodule $M$ and a $(B, C)$-bimodule $N$, in that order, is $M \otimes_B N$. Adopting the usual composition convention switches the meaning of "left" and "right," but this convention makes things work out very nicely as follows.

Proposition: An $(A, B)$-bimodule $M$ has a left adjoint iff it is finitely presented projective as a left $A$-module, in which case its left adjoint is $\text{Hom}_A(M, A)$. It has a right adjoint iff it is finitely presented projective as a right $B$-module, in which case its right adjoint is $\text{Hom}_B(M, B)$.

KV is the special case where we only consider algebras of the form $k^n$ for $k$ a field and $(k^n, k^m)$-bimodules, which concretely can be thought of as $n \times m$ matrices of $k$-vector spaces, where all of the vector spaces are finite-dimensional. These are all finitely presented projective on both sides, and so we conclude that every morphism of KV 2-vector spaces has both a left and a right adjoint, both of which I believe are concretely given by sending a matrix $V_{ij}$ of vector spaces to the transposed and dualized matrix $V_{ji}^{\ast}$, but I haven't checked this in detail.

(Note that because I'm thinking about 1-morphisms as matrices of vector spaces instead of taking a skeleton and identifying vector spaces with their dimensions, when I write down taking the dual I'm making some implicit claims about functoriality and 2-morphisms that I don't need to keep explicit track of. It's actually more work to work with the skeleton because then I need to distinguish a vector space from its dual by talking a lot more explicitly about 2-morphisms.)

As in the previous question, I continue to suggest that everything is much cleaner if you think in terms of bimodules. So we'll ask the more general question: in the 2-category $\text{Bim}(k)$ of $k$-algebras, $k$-bimodules, and bimodule homomorphisms, what is an adjoint pair?

The answer is worked out in this blog post, although I'll warn you that in the convention of that blog post composition of 1-morphisms is written in diagrammatic order, which means concretely that the composition of an $(A, B)$-bimodule $M$ and a $(B, C)$-bimodule $N$, in that order, is $M \otimes_B N$. Adopting the usual composition convention switches the meaning of "left" and "right," but this convention makes things work out very nicely as follows.

Proposition: An $(A, B)$-bimodule $M$ has a left adjoint iff it is finitely presented projective as a left $A$-module, in which case its left adjoint is $\text{Hom}_A(M, A)$. It has a right adjoint iff it is finitely presented projective as a right $B$-module, in which case its right adjoint is $\text{Hom}_B(M, B)$.

KV is the special case where we only consider algebras of the form $k^n$ for $k$ a field and $(k^n, k^m)$-bimodules, which concretely can be thought of as $n \times m$ matrices of $k$-vector spaces, where all of the vector spaces are finite-dimensional. These are all finitely presented projective on both sides, and so we conclude that every morphism of KV 2-vector spaces has both a left and a right adjoint, both of which I believe are concretely given by sending a matrix $V_{ij}$ of vector spaces to the transposed and dualized matrix $V_{ji}^{\ast}$, but I haven't checked this in detail.

(Note that because I'm thinking about 1-morphisms as matrices of vector spaces instead of taking a skeleton and identifying vector spaces with their dimensions, when I write down taking the dual I'm making some implicit claims about functoriality and 2-morphisms that I don't need to keep explicit track of. It's actually more work to work with the skeleton because then I need to distinguish a vector space from its dual by talking a lot more explicitly about 2-morphisms, and I sort of have to conjure the 2-morphisms out of thin air rather than letting them naturally arise out of e.g. the dual pairing.)

Source Link
Qiaochu Yuan
  • 118.2k
  • 40
  • 447
  • 741

As in the previous question, I continue to suggest that everything is much cleaner if you think in terms of bimodules. So we'll ask the more general question: in the 2-category $\text{Bim}(k)$ of $k$-algebras, $k$-bimodules, and bimodule homomorphisms, what is an adjoint pair?

The answer is worked out in this blog post, although I'll warn you that in the convention of that blog post composition of 1-morphisms is written in diagrammatic order, which means concretely that the composition of an $(A, B)$-bimodule $M$ and a $(B, C)$-bimodule $N$, in that order, is $M \otimes_B N$. Adopting the usual composition convention switches the meaning of "left" and "right," but this convention makes things work out very nicely as follows.

Proposition: An $(A, B)$-bimodule $M$ has a left adjoint iff it is finitely presented projective as a left $A$-module, in which case its left adjoint is $\text{Hom}_A(M, A)$. It has a right adjoint iff it is finitely presented projective as a right $B$-module, in which case its right adjoint is $\text{Hom}_B(M, B)$.

KV is the special case where we only consider algebras of the form $k^n$ for $k$ a field and $(k^n, k^m)$-bimodules, which concretely can be thought of as $n \times m$ matrices of $k$-vector spaces, where all of the vector spaces are finite-dimensional. These are all finitely presented projective on both sides, and so we conclude that every morphism of KV 2-vector spaces has both a left and a right adjoint, both of which I believe are concretely given by sending a matrix $V_{ij}$ of vector spaces to the transposed and dualized matrix $V_{ji}^{\ast}$, but I haven't checked this in detail.

(Note that because I'm thinking about 1-morphisms as matrices of vector spaces instead of taking a skeleton and identifying vector spaces with their dimensions, when I write down taking the dual I'm making some implicit claims about functoriality and 2-morphisms that I don't need to keep explicit track of. It's actually more work to work with the skeleton because then I need to distinguish a vector space from its dual by talking a lot more explicitly about 2-morphisms.)