Revisions to What's the correct notion of determinant of a bilinear pairing?

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edited Apr 12, 2017 at 7:22

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By a pairing on a vector space $V$, I mean a linear map $A : V \otimes V \to R$. If $V$ is $n$-dimensional ($n < \infty$), then I can define the determinant of $A$ by considering the canonical action of $A$ on $\bigwedge^n V \otimes \bigwedge^n V \to R$. In fact, I should define $\det A$ to be this map, which I will call $\bigwedge^n A$. Remember that $\bigwedge^n V$ is one-dimensional, and pairs canonically with the space of volume forms on $V$. So if $V$ has a volume form $\mathrm{vol}$, we can define $\det A$ as $\bigwedge^n A(\mathrm{vol}^{-1} \otimes \mathrm{vol}^{-1})$.

For comparison, if I had an operator $A: V\to V$, then I would look at the map $\bigwedge^n A: \bigwedge^n V \to \bigwedge^n V$. But $\bigwedge^n$ is one-dimensional, so $\bigwedge^n A$ consists of multiplication by a scalar, and this scalar is $\det A$. On the other hand, if you pick an identification $V \cong V^*$, then you can think of a pairing as an operator; the identification determines a non-zero volume form, and the notions of determinants are the same.

In any case, these definitions don't work for infinite-dimensional vector spaces, because there's no ``wedge-top''. I'd like a notion of "determinant of a pairing" like the zeta-function regularized determinant of an operator zeta-function regularized determinant of an operator.

By a pairing on a vector space $V$, I mean a linear map $A : V \otimes V \to R$. If $V$ is $n$-dimensional ($n < \infty$), then I can define the determinant of $A$ by considering the canonical action of $A$ on $\bigwedge^n V \otimes \bigwedge^n V \to R$. In fact, I should define $\det A$ to be this map, which I will call $\bigwedge^n A$. Remember that $\bigwedge^n V$ is one-dimensional, and pairs canonically with the space of volume forms on $V$. So if $V$ has a volume form $\mathrm{vol}$, we can define $\det A$ as $\bigwedge^n A(\mathrm{vol}^{-1} \otimes \mathrm{vol}^{-1})$.

For comparison, if I had an operator $A: V\to V$, then I would look at the map $\bigwedge^n A: \bigwedge^n V \to \bigwedge^n V$. But $\bigwedge^n$ is one-dimensional, so $\bigwedge^n A$ consists of multiplication by a scalar, and this scalar is $\det A$. On the other hand, if you pick an identification $V \cong V^*$, then you can think of a pairing as an operator; the identification determines a non-zero volume form, and the notions of determinants are the same.

In any case, these definitions don't work for infinite-dimensional vector spaces, because there's no ``wedge-top''. I'd like a notion of "determinant of a pairing" like the zeta-function regularized determinant of an operator.

By a pairing on a vector space $V$, I mean a linear map $A : V \otimes V \to R$. If $V$ is $n$-dimensional ($n < \infty$), then I can define the determinant of $A$ by considering the canonical action of $A$ on $\bigwedge^n V \otimes \bigwedge^n V \to R$. In fact, I should define $\det A$ to be this map, which I will call $\bigwedge^n A$. Remember that $\bigwedge^n V$ is one-dimensional, and pairs canonically with the space of volume forms on $V$. So if $V$ has a volume form $\mathrm{vol}$, we can define $\det A$ as $\bigwedge^n A(\mathrm{vol}^{-1} \otimes \mathrm{vol}^{-1})$.

For comparison, if I had an operator $A: V\to V$, then I would look at the map $\bigwedge^n A: \bigwedge^n V \to \bigwedge^n V$. But $\bigwedge^n$ is one-dimensional, so $\bigwedge^n A$ consists of multiplication by a scalar, and this scalar is $\det A$. On the other hand, if you pick an identification $V \cong V^*$, then you can think of a pairing as an operator; the identification determines a non-zero volume form, and the notions of determinants are the same.

In any case, these definitions don't work for infinite-dimensional vector spaces, because there's no ``wedge-top''. I'd like a notion of "determinant of a pairing" like the zeta-function regularized determinant of an operator.

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edited Jul 31, 2013 at 8:27

Ricardo Andrade

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By a pairing on a vector space $V$, I mean a linear map $A : V \otimes V \to R$. If $V$ is $n$-dimensional ($n < \infty$$n < \infty$), then I can define the determinant of $A$ by considering the canonical action of $A$ on $\bigwedge^n V \otimes \bigwedge^n V \to R$. In fact, I should define $\det A$ to be this map, which I will call $\bigwedge^n A$. Remember that $\bigwedge^n V$ is one-dimensional, and pairs canonically with the space of volume forms on $V$. So if $V$ has a volume form $\vol$$\mathrm{vol}$, we can define $\det A$ as $\bigwedge^n A(\vol^{-1} \otimes \vol^{-1})$$\bigwedge^n A(\mathrm{vol}^{-1} \otimes \mathrm{vol}^{-1})$.

For comparison, if I had an operator $A: V\to V$, then I would look at the map $\bigwedge^n A: \bigwedge^n V \to \bigwedge^n V$. But $\bigwedge^n$ is one-dimensional, so $\bigwedge^n A$ consists of multiplication by a scalar, and this scalar is $\det A$. On the other hand, if you pick an identification $V \cong V^*$, then you can think of a pairing as an operator; the identification determines a non-zero volume form, and the notions of determinants are the same.

In any case, these definitions don't work for infinite-dimensional vector spaces, because there's no ``wedge-top''. I'd like a notion of "determinant of a pairing" like the zeta-function regularized determinant of an operator.

By a pairing on a vector space $V$, I mean a linear map $A : V \otimes V \to R$. If $V$ is $n$-dimensional ($n < \infty$), then I can define the determinant of $A$ by considering the canonical action of $A$ on $\bigwedge^n V \otimes \bigwedge^n V \to R$. In fact, I should define $\det A$ to be this map, which I will call $\bigwedge^n A$. Remember that $\bigwedge^n V$ is one-dimensional, and pairs canonically with the space of volume forms on $V$. So if $V$ has a volume form $\vol$, we can define $\det A$ as $\bigwedge^n A(\vol^{-1} \otimes \vol^{-1})$.

For comparison, if I had an operator $A: V\to V$, then I would look at the map $\bigwedge^n A: \bigwedge^n V \to \bigwedge^n V$. But $\bigwedge^n$ is one-dimensional, so $\bigwedge^n A$ consists of multiplication by a scalar, and this scalar is $\det A$. On the other hand, if you pick an identification $V \cong V^*$, then you can think of a pairing as an operator; the identification determines a non-zero volume form, and the notions of determinants are the same.

In any case, these definitions don't work for infinite-dimensional vector spaces, because there's no ``wedge-top''. I'd like a notion of "determinant of a pairing" like the zeta-function regularized determinant of an operator.

By a pairing on a vector space $V$, I mean a linear map $A : V \otimes V \to R$. If $V$ is $n$-dimensional ($n < \infty$), then I can define the determinant of $A$ by considering the canonical action of $A$ on $\bigwedge^n V \otimes \bigwedge^n V \to R$. In fact, I should define $\det A$ to be this map, which I will call $\bigwedge^n A$. Remember that $\bigwedge^n V$ is one-dimensional, and pairs canonically with the space of volume forms on $V$. So if $V$ has a volume form $\mathrm{vol}$, we can define $\det A$ as $\bigwedge^n A(\mathrm{vol}^{-1} \otimes \mathrm{vol}^{-1})$.

For comparison, if I had an operator $A: V\to V$, then I would look at the map $\bigwedge^n A: \bigwedge^n V \to \bigwedge^n V$. But $\bigwedge^n$ is one-dimensional, so $\bigwedge^n A$ consists of multiplication by a scalar, and this scalar is $\det A$. On the other hand, if you pick an identification $V \cong V^*$, then you can think of a pairing as an operator; the identification determines a non-zero volume form, and the notions of determinants are the same.

In any case, these definitions don't work for infinite-dimensional vector spaces, because there's no ``wedge-top''. I'd like a notion of "determinant of a pairing" like the zeta-function regularized determinant of an operator.