The product formula for regularized Fredholm determinants
Authors: Britz Thomas , Carey Alan , Gesztesy Fritz , Nichols Roger , Sukochev Fedor , Zanin Dmitriy 
Published: February 2021

For trace class operators

A
,
B
∈

B

1

(

H

)

A, B \in \mathcal {B}_1(\mathcal {H})

(

H

\mathcal {H}

a complex, separable Hilbert space), the product formula for Fredholm determinants holds in the familiar form
\[
<mml:math xmlns:mml="http://www.w3.org/1998/Math/MathML" alttext="det Subscript script upper H Baseline left-parenthesis left-parenthesis upper I Subscript script upper H Baseline minus upper A right-parenthesis left-parenthesis upper I Subscript script upper H Baseline minus upper B right-parenthesis right-parenthesis equals det Subscript script upper H Baseline left-parenthesis upper I Subscript script upper H Baseline minus upper A right-parenthesis det Subscript script upper H Baseline left-parenthesis upper I Subscript script upper H Baseline minus upper B right-parenthesis period">

det

H

(
(

I

H

−
A
)
(

I

H

−
B
)
)
=

det

H

(

I

H

−
A
)

det

H

(

I

H

−
B
)
.

{\det }_{\mathcal {H}} ((I_{\mathcal {H}} - A) (I_{\mathcal {H}} - B)) = {\det }_{\mathcal {H}} (I_{\mathcal {H}} - A) {\det }_{\mathcal {H}} (I_{\mathcal {H}} - B).

\]
When trace class operators are replaced by Hilbert–Schmidt operators

A
,
B
∈

B

2

(

H

)

A, B \in \mathcal {B}_2(\mathcal {H})

and the Fredholm determinant

det

H

(

I

H

−
A
)

{\det }_{\mathcal {H}}(I_{\mathcal {H}} - A)

,

A
∈

B

1

(

H

)

A \in \mathcal {B}_1(\mathcal {H})

, by the 2nd regularized Fredholm determinant
<mml:math xmlns:mml="http://www.w3.org/1998/Math/MathML" alttext="det Subscript script upper H comma 2 Baseline left-parenthesis upper I Subscript script upper H Baseline minus upper A right-parenthesis equals det Subscript script upper H Baseline left-parenthesis left-parenthesis upper I Subscript script upper H Baseline minus upper A right-parenthesis exp left-parenthesis upper A right-parenthesis right-parenthesis">

det

H

,
2

(

I

H

−
A
)
=

det

H

(
(

I

H

−
A
)
exp
⁡
(
A
)
)

{\det }_{\mathcal {H},2}(I_{\mathcal {H}} - A) = {\det }_{\mathcal {H}} ((I_{\mathcal {H}} - A) \exp (A))

,

A
∈

B

2

(

H

)

A \in \mathcal {B}_2(\mathcal {H})

, the product formula must be replaced by
<mml:math xmlns:mml="http://www.w3.org/1998/Math/MathML" alttext="StartLayout 1st Row 1st Column det Subscript script upper H comma 2 Baseline left-parenthesis left-parenthesis upper I Subscript script upper H Baseline minus upper A right-parenthesis left-parenthesis upper I Subscript script upper H Baseline minus upper B right-parenthesis right-parenthesis 2nd Column a m p semicolon equals det Subscript script upper H comma 2 Baseline left-parenthesis upper I Subscript script upper H Baseline minus upper A right-parenthesis det Subscript script upper H comma 2 Baseline left-parenthesis upper I Subscript script upper H Baseline minus upper B right-parenthesis 2nd Row 1st Column Blank 2nd Column a m p semicolon times exp left-parenthesis minus trace Subscript script upper H Baseline left-parenthesis upper A upper B right-parenthesis right-parenthesis period EndLayout">

det

H

,
2

(
(

I

H

−
A
)
(

I

H

−
B
)
)

a
m
p
;
=

det

H

,
2

(

I

H

−
A
)

det

H

,
2

(

I

H

−
B
)

a
m
p
;

×
exp
⁡
(
−

tr

H

⁡
(
A
B
)
)
.

\begin{align*} {\det }_{\mathcal {H},2} ((I_{\mathcal {H}} - A) (I_{\mathcal {H}} - B)) &= {\det }_{\mathcal {H},2} (I_{\mathcal {H}} - A) {\det }_{\mathcal {H},2} (I_{\mathcal {H}} - B) \\ & \quad \times \exp (- \operatorname {tr}_{\mathcal {H}}(AB)). \end{align*}

The product formula for the case of higher regularized Fredholm determinants

det

H

,
k

(

I

H

−
A
)

{\det }_{\mathcal {H},k}(I_{\mathcal {H}} - A)

,

A
∈

B

k

(

H

)

A \in \mathcal {B}_k(\mathcal {H})

,

k
∈

N

k \in \mathbb {N}

,

k
⩾
2

k \geqslant 2

, does not seem to be easily accessible and hence this note aims at filling this gap in the literature.