Fredholm determinant

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Let ${\displaystyle H}$ be a Hilbert space and ${\displaystyle G}$ the set of bounded invertible operators on ${\displaystyle H}$ of the form ${\displaystyle I+T}$, where ${\displaystyle T}$ is a trace-class operator. ${\displaystyle G}$ is a group because

• The set of trace-class operators is an ideal in the algebra of bounded linear operators, so ${\displaystyle (I+T)(I+T')-I=T+T'+TT'}$ is trace-class.
• ${\textstyle (I+T)^{-1}-I=-T(I+T)^{-1},}$ so ${\displaystyle (I+T)^{-1}-I}$ is trace class if ${\displaystyle T}$ is.

${\displaystyle G}$ has a natural metric given by ${\displaystyle d(X,Y)=\|X-Y\|_{1}}$, where ${\displaystyle \|X\|_{1}=\sum _{i}|\lambda _{i}(X)|}$ is the trace-class norm.

One definition uses the exponential trace formula. For finite-dimensional matrices, we have ${\textstyle \det(I+A)=e^{\operatorname {Tr} (\ln(I+A))}}$, which expands in Taylor series to$${\displaystyle \operatorname {det} (I+A)=\exp \left(\sum _{n=1}^{\infty }{\frac {(-1)^{n+1}}{n}}\operatorname {Tr} \left(A^{n}\right)\right)}$$This then generalizes directly to trace-class operators.

In the finite-dimensional case, the determinant of an operator can be interpreted as the factor by which it scales the (oriented) volume of a parallelepiped. This can be generalized to infinite dimensions.

In finite dimensions, by expanding the definition of determinant as a sum over permutations,$${\displaystyle \det(I+A)=\sum _{S}\det(A_{SS})}$$where ${\displaystyle S}$ ranges over all subsets of the index set of ${\displaystyle A}$. For example, when the index set is ${\displaystyle \{1,2\}}$ then ${\displaystyle S=\{\},\{1\},\{2\},\{1,2\}}$.

If ${\displaystyle H}$ is an ${\displaystyle n}$-dimensional Hilbert space with inner product ${\displaystyle (\cdot ,\cdot )}$, then the ${\displaystyle k}$-th exterior power ${\displaystyle \Lambda ^{k}H}$ is also a ${\displaystyle {\binom {n}{k}}}$-dimensional Hilbert space, with inner product $${\displaystyle (v_{1}\wedge v_{2}\wedge \cdots \wedge v_{k},w_{1}\wedge w_{2}\wedge \cdots \wedge w_{k})=\det(v_{i},w_{j}).}$$In particular $${\displaystyle e_{i_{1}}\wedge e_{i_{2}}\wedge \cdots \wedge e_{i_{k}},\qquad (i_{1}<i_{2}<\cdots <i_{k})}$$gives an orthonormal basis of ${\displaystyle \Lambda ^{k}H}$ if ${\displaystyle (e_{i})}$ is an orthonormal basis of ${\displaystyle H}$.

If ${\displaystyle A}$ is an operator on ${\displaystyle H}$, then ${\displaystyle A}$ functorially defines a bounded operator ${\displaystyle \Lambda ^{k}(A)}$ on ${\displaystyle \Lambda ^{k}H}$ by $${\displaystyle \Lambda ^{k}(A)v_{1}\wedge v_{2}\wedge \cdots \wedge v_{k}=Av_{1}\wedge Av_{2}\wedge \cdots \wedge Av_{k}.}$$By definition of trace, we have$${\displaystyle \operatorname {Tr} \left(\Lambda ^{k}A\right)=\sum _{1\leq i_{1}<\cdots <i_{k}\leq n}(e_{i_{1}}\wedge e_{i_{2}}\wedge \cdots \wedge e_{i_{k}},Ae_{i_{1}}\wedge Ae_{i_{2}}\wedge \cdots \wedge Ae_{i_{k}})}$$The summand simplifies to ${\displaystyle \det[(e_{i_{j}},Ae_{i_{j'}})]=\det(A_{SS})}$ where ${\displaystyle S=\{i_{1},\dots ,i_{k}\}}$. Thus $${\displaystyle {\begin{aligned}&\operatorname {Tr} \Lambda ^{k}(A)=\sum _{|S|=k}\operatorname {det} \left(A_{SS}\right).\\&\operatorname {det} (I+A)=\sum _{k=0}^{n}\operatorname {Tr} \Lambda ^{k}(A).\end{aligned}}}$$This generalizes to infinite-dimensional Hilbert spaces, and bounded trace-class operators, allowing us to define the Fredholm determinant by$${\displaystyle \det(I+A)=\sum _{k=0}^{\infty }\operatorname {Tr} \Lambda ^{k}(A)}$$To show that the definition makes sense, note that if ${\displaystyle A}$ is trace-class, then ${\displaystyle \Lambda ^{k}(A)}$ is also trace-class with ${\textstyle \|\Lambda ^{k}(A)\|_{1}\leq \|A\|_{1}^{k}/k!}$, thus ${\displaystyle \sum _{k=0}^{\infty }|\operatorname {Tr} \Lambda ^{k}(A)|\leq e^{\|A\|_{1}}}$.

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By default, all operators are assumed trace-class.

• ${\textstyle \det(I+A)\cdot \det(I+B)=\det(I+A)(I+B).}$
• ${\textstyle z\mapsto \det(I+zA)=\sum _{k=0}^{\infty }z^{k}\operatorname {Tr} \Lambda ^{k}(A)}$ defines an entire function, with ${\textstyle \left|\det(I+zA)\right|\leq \exp(|z|\cdot \|A\|_{1}).}$

• The function ${\displaystyle A\mapsto \det(I+A)}$ is continuous on trace-class operators, with

$${\displaystyle \left|\det(I+A)-\det(I+B)\right|\leq \|A-B\|_{1}\exp(\|A\|_{1}+\|B\|_{1}+1).}$$ One can improve this inequality slightly to the following, as noted in (Simon 2005, Chapter 5): $${\displaystyle \left|\det(I+A)-\det(I+B)\right|\leq \|A-B\|_{1}\exp(\max(\|A\|_{1},\|B\|_{1})+1).}$$

• The function ${\displaystyle \det }$ defines a homomorphism of type ${\displaystyle G\to \mathbb {C} ^{\times }}$ where ${\displaystyle \mathbb {C} ^{\times }}$ the multiplicative group of nonzero complex numbers (since elements of ${\displaystyle G}$ are invertible).
• If ${\displaystyle T}$ is in ${\displaystyle G}$ and ${\displaystyle X}$ is invertible, ${\textstyle \det XTX^{-1}=\det T.}$

• ${\textstyle \det e^{A}=\exp \,\operatorname {Tr} (A).}$

• ${\textstyle \log \det(I+zA)=\operatorname {Tr} (\log {(I+zA)})=\sum _{k=1}^{\infty }(-1)^{k+1}{\frac {\operatorname {Tr} A^{k}}{k}}z^{k}}$

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The Fredholm determinant is often applied to integral operators. Let the trace-class operator ${\displaystyle T}$ be an integral operator given by a kernel ${\displaystyle K(x,y)}$, then the Fredholm determinant is defined, like before, by$${\displaystyle \det(I-\lambda T)=\sum _{n=0}^{\infty }(-\lambda )^{n}\operatorname {Tr} \Lambda ^{n}(T)=\exp {\left(-\sum _{n=1}^{\infty }{\frac {\operatorname {Tr} (T^{n})}{n}}\lambda ^{n}\right)}}$$where ${\displaystyle T}$ is an integral operator. The trace of the operator ${\displaystyle T}$ and its alternating powers is given in terms of the kernel ${\displaystyle K}$ by $${\displaystyle \operatorname {Tr} T=\int K(x,x)\,dx}$$ and $${\displaystyle \operatorname {Tr} \Lambda ^{2}(T)={\frac {1}{2!}}\iint \left(K(x,x)K(y,y)-K(x,y)K(y,x)\right)dx\,dy}$$ and in general $${\displaystyle \operatorname {Tr} \Lambda ^{n}(T)={\frac {1}{n!}}\int \cdots \int \det[K(x_{i},x_{j})]_{i,j\in 1:n}\,dx_{1:n}}$$The trace is well-defined for these kernels, since these are trace-class or nuclear operators.

To see that this is a special case of the previous section's general definition, note that,$${\displaystyle \operatorname {Tr} \left(\Lambda ^{k}A\right)=\sum _{1\leq i_{1}<\cdots <i_{k}\leq n}(e_{i_{1}}\wedge e_{i_{2}}\wedge \cdots \wedge e_{i_{k}},Ae_{i_{1}}\wedge Ae_{i_{2}}\wedge \cdots \wedge Ae_{i_{k}})}$$is equivalent to$${\displaystyle {\frac {1}{k!}}\sum _{i_{1},\cdots ,i_{k}\in 1:n,{\text{ all different}}}\det(A_{SS})}$$where ${\displaystyle S}$ is the ordered sequence ${\displaystyle i_{1},\dots ,i_{k}}$. Now, to convert this to integral equations, a matrix becomes a kernel, and a summation over indices becomes an integral over coordinates.

The above argument is intuitive. A proper definition requires a presentation showing that each of the manipulations are well-defined, convergent, and so on, for the given situation for which the Fredholm determinant is contemplated. Since the kernel ${\displaystyle K}$ may be defined for a large variety of Hilbert spaces and Banach spaces, this is a non-trivial exercise.

The original (Fredholm 1903) considered the integral equation$${\displaystyle u(x)+z\int _{a}^{b}K(x,y)u(y)dy=f(x)\quad (x\in (a,b))}$$which can be written as ${\displaystyle (I+zA)u=f}$. Fredholm proved that this equation has a unique solution iff ${\displaystyle \det(I+zA)\neq 0}$.

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A function ${\displaystyle F(t)}$ from ${\displaystyle (a,b)}$ into ${\displaystyle G}$ is said to be differentiable if ${\displaystyle F(t)-I}$ is differentiable as a map into the trace-class operators, i.e. if the limit

$${\displaystyle {\dot {F}}(t)=\lim _{h\to 0}{F(t+h)-F(t) \over h}}$$

exists in trace-class norm.

If ${\displaystyle g(t)}$ is a differentiable function with values in trace-class operators, then so too is ${\displaystyle \exp g(t)}$ and

$${\displaystyle F^{-1}{\dot {F}}={\operatorname {id} -\exp -\operatorname {ad} g(t) \over \operatorname {ad} g(t)}\cdot {\dot {g}}(t),}$$

where $${\displaystyle \operatorname {ad} (X)\cdot Y=XY-YX.}$$

Israel Gohberg and Mark Krein proved that if ${\displaystyle F}$ is a differentiable function into ${\displaystyle G}$, then ${\displaystyle f=\det F}$ is a differentiable map into ${\displaystyle \mathbb {C} ^{*}}$ with $${\displaystyle f^{-1}{\dot {f}}=\operatorname {Tr} F^{-1}{\dot {F}}.}$$

This result was used by Joel Pincus, William Helton and Roger Howe to prove that if ${\displaystyle A}$ and ${\displaystyle B}$ are bounded operators with trace-class commutator ${\displaystyle AB-BA}$, then

$${\displaystyle \det e^{A}e^{B}e^{-A}e^{-B}=\exp \operatorname {Tr} (AB-BA).}$$