Trace Ideals and Fredholm Determinants

In infinite dimensions, a displayed diagonal sum or eigenvalue product is not automatically a trace or determinant. The canonical threshold is summability in trace norm: a trace-class operator $T\in\mathcal S_1(H)$ has a finite, basis-independent operator trace, and an identity-plus-trace-class operator $I+K$ has an ordinary Fredholm determinant $\det_{\!F}(I+K)$. Compactness or Hilbert–Schmidt membership alone is not enough.

This distinction is especially important in QFT. A formal expression such as $\operatorname{Tr}\log A$ does not prove that $\det A$ exists. A determinant ratio is controlled here only after it has been reduced to $\det_{\!F}(I+K)$ with a proved $K\in\mathcal S_1$. A one-dimensional Euclidean fluctuation problem below meets that test and yields an exact ratio. Its higher-dimensional analogue fails the test, identifying the point at which a modified determinant, heat-kernel or zeta prescription, and ultimately renormalization become additional structure.

Required background. Bounded, Compact, and Integral Operators supplies compactness, singular values, Hilbert–Schmidt operators, square-integrable kernels, and finite-rank approximation.

Trace-class operators and ordinary Fredholm determinants

This page develops the hypothesis test for Schatten ideals, the basis-independent trace, the ordinary Fredholm determinant, and a method for validating finite-rank approximations in trace norm. It does not develop general regularized determinants, functional integration over physical fields, one-loop matching, or production numerical software.

Its QFT-facing application is a mathematically controlled determinant ratio before zeta or heat-kernel regularization. The physical interpretation remains with Renormalization and Effective Field Theory.

Throughout, $H$ is a complex separable Hilbert space, the bra is conjugate-linear, and $T^\dagger$ denotes the Hilbert-space adjoint. The worked fluctuation problem is explicitly Euclidean. The notation $\operatorname{Tr}_H$ denotes an operator trace; $\operatorname{tr}$ is reserved for an ordinary finite-dimensional matrix trace when a distinction is needed.

Singular values measure the required summability

For a compact operator $K$, let

$$|K|=(K^\dagger K)^{1/2}.$$

The eigenvalues of $|K|$, repeated according to multiplicity and arranged in nonincreasing order, are the singular values $s_1(K),s_2(K),\ldots$. For $1\leq p<\infty$, the Schatten class is

$$\mathcal S_p(H) = \left\{ K\in\mathcal K(H): \sum_{n=1}^{\infty}s_n(K)^p<\infty \right\},$$

with norm

$$\|K\|_p = \left(\sum_{n=1}^{\infty}s_n(K)^p\right)^{1/p}.$$

On an infinite-dimensional $H$, the relevant strict inclusions are

$$\mathcal S_1(H) \subsetneq \mathcal S_2(H) \subsetneq \mathcal K(H) \subsetneq \mathcal B(H).$$

Here $\mathcal S_1$ is the trace class and $\mathcal S_2$ is the Hilbert–Schmidt class. Each $\mathcal S_p$ is a two-sided ideal: if $K\in\mathcal S_p$ and $A,B\in\mathcal B(H)$, then

$$AKB\in\mathcal S_p, \qquad \|AKB\|_p \leq \|A\|\,\|K\|_p\,\|B\|.$$

Finite-rank operators are dense in $\mathcal S_p$ in the $p$-norm. The Schatten Hölder inequality gives, in particular,

$$A,B\in\mathcal S_2 \quad\Longrightarrow\quad AB\in\mathcal S_1, \qquad \|AB\|_1\leq\|A\|_2\|B\|_2.$$

Conversely, every trace-class operator factors as a product of two Hilbert–Schmidt operators. Indeed, the polar decomposition $T=U|T|$ gives

$$T=(U|T|^{1/2})|T|^{1/2},$$

and

$$\left\||T|^{1/2}\right\|_2^2 = \left\|U|T|^{1/2}\right\|_2^2 = \sum_n s_n(T) <\infty.$$

Factorization is often a much easier trace-class test than estimating singular values directly. These ideal, inclusion, and factorization results are developed in Kostenko 2019, §§3.1–3.3, PDF and Simon 2005, Chapter 2.

A diagonal model separates the classes

Let $(e_n)_{n\geq1}$ be an orthonormal basis and $D_a e_n=a_ne_n$, where $a_n\to0$. Then

$$D_a\in\mathcal S_p \quad\Longleftrightarrow\quad (a_n)\in\ell^p.$$

For the positive diagonal operator $D e_n=n^{-1}e_n$,

$$\sum_{n=1}^{\infty}s_n(D)^2 = \sum_{n=1}^{\infty}\frac1{n^2} <\infty, \qquad \sum_{n=1}^{\infty}s_n(D) = \sum_{n=1}^{\infty}\frac1n =\infty.$$

Thus $D$ is compact and Hilbert–Schmidt but not trace class. If $P_N$ projects onto $\operatorname{span}\{e_1,\ldots,e_N\}$, then

$$\|P_NDP_N-D\|\longrightarrow0, \qquad \|P_NDP_N-D\|_2\longrightarrow0,$$

while

$$\operatorname{tr}(P_NDP_N) = \sum_{n=1}^{N}\frac1n\longrightarrow\infty$$

and

$$\det(I+P_NDP_N) = \prod_{n=1}^{N}\left(1+\frac1n\right) =N+1.$$

Operator-norm convergence—and even Hilbert–Schmidt convergence—therefore does not control traces or determinants. Trace norm is the topology that does.

Trace class makes the diagonal sum intrinsic

For $T\in\mathcal S_1(H)$ and any orthonormal basis $(e_n)$, define

$$\operatorname{Tr}_H T = \sum_{n=1}^{\infty} \langle e_n|Te_n\rangle.$$

The series is absolutely convergent, its value is independent of the basis, and

$$\left|\operatorname{Tr}_H T\right| \leq \|T\|_1.$$

For $T\geq0$, $\operatorname{Tr}_H T=\|T\|_1=\sum_n s_n(T)$. The adjoint and cyclicity rules take the controlled forms

$$\operatorname{Tr}_H(T^\dagger) = \overline{\operatorname{Tr}_H T},$$

and, for $B\in\mathcal B(H)$,

$$\operatorname{Tr}_H(BT) = \operatorname{Tr}_H(TB).$$

The latter statement works because both products are trace class. It does not license cyclic rearrangement of arbitrary unbounded products. If $A,B\in\mathcal S_2$, the product theorem supplies the needed hypothesis, so

$$\operatorname{Tr}_H(AB) = \operatorname{Tr}_H(BA), \qquad \left|\operatorname{Tr}_H(AB)\right| \leq \|A\|_2\|B\|_2.$$

For a rank-one operator,

$$\operatorname{Tr}_H\!\left(|u\rangle\langle v|\right) = \langle v|u\rangle.$$

Lidskii’s theorem is the nontrivial bridge from the basis definition to the spectrum. If $\lambda_j(T)$ are the nonzero eigenvalues of a trace-class operator, counted with algebraic multiplicity, then

$$\operatorname{Tr}_H T = \sum_j\lambda_j(T).$$

This is a theorem, not a definition for arbitrary compact operators. For example, the diagonal series of $D_{(-1)^n/n}$ converges conditionally in its displayed order, but the operator is not trace class. Reordering the basis can change the series, so that number is not an operator trace. See Kostenko 2019, §3.2 and Theorem 3.4.7, PDF and Simon 2005, Chapter 3 for the basis-independent trace, cyclicity, and Lidskii theorem.

A square-integrable kernel does not have a traceable diagonal

On a finite-measure domain $X$, a kernel $k\in L^2(X\times X)$ defines a Hilbert–Schmidt integral operator. This proves membership in $\mathcal S_2$, not in $\mathcal S_1$. Moreover, an $L^2$ kernel is an almost-everywhere equivalence class. On a nonatomic space its values on the diagonal can be changed without changing the operator, so

$$\int_X k(x,x)\,\mathrm dx$$

need not even be well defined.

A factorization into two Hilbert–Schmidt operators proves trace class. Under stronger Mercer-type hypotheses—such as a continuous positive-definite kernel on a compact metric measure space—the diagonal integral can agree with the operator trace. Neither conclusion follows from $L^2$ membership or an undeclared pointwise representative alone (Kostenko 2019, §3.3, PDF; Bornemann 2010, §2).

The ordinary Fredholm determinant

Let $K\in\mathcal S_1(H)$. A definition that does not assume normality is

$$\det_{\!F}(I+K) = \sum_{r=0}^{\infty} \operatorname{Tr}_{\wedge^r H}\!\left(\wedge^r K\right),$$

where the $r=0$ term is $1$. The series converges absolutely. Equivalently, if finite-rank operators $K_N$ satisfy $\|K_N-K\|_1\to0$, then

$$\det_{\!F}(I+K) = \lim_{N\to\infty}\det(I+K_N).$$

The finite-dimensional determinant is taken on any finite-dimensional subspace containing the effective range of $K_N$; extending it by the identity does not change the value.

Lidskii’s theorem and exterior algebra give the eigenvalue product

$$\det_{\!F}(I+K) = \prod_j\left(1+\lambda_j(K)\right),$$

with algebraic multiplicities. The product contains eigenvalues, not singular values. Its absolute convergence is guaranteed by trace class.

Three properties make this determinant useful:

1. Invertibility test. $\det_{\!F}(I+K)=0$ exactly when $I+K$ is not invertible.

2. Multiplicativity. If $A,B\in\mathcal S_1$, then

   $$\det_{\!F}\!\left((I+A)(I+B)\right) = \det_{\!F}(I+A)\det_{\!F}(I+B).$$

3. Trace-norm continuity. One convenient estimate is

   $$\begin{aligned} &\left| \det_{\!F}(I+A)-\det_{\!F}(I+B) \right|\\ &\qquad\leq \|A-B\|_1 \exp\!\left( 1+\|A\|_1+\|B\|_1 \right). \end{aligned}$$

   Thus a certified trace-norm error gives a certified determinant error. This estimate is Eq. (3.4.19) in Kostenko 2019, §3.4, PDF; Bornemann 2010, §4 gives an independent perturbation treatment.

The rank-one identity

$$\det_{\!F}\!\left(I+|u\rangle\langle v|\right) = 1+\langle v|u\rangle$$

is a useful normalization check.

The equivalent definitions, eigenvalue product, invertibility criterion, and multiplicativity are collected in Kostenko 2019, §§3.4–3.5, PDF and Bornemann 2010, §3.

The logarithm is local

The determinant is a globally defined scalar. Its logarithm is not. If $|z|\,\|K\|<1$, the branch fixed to vanish at $z=0$ obeys

$$\log\det_{\!F}(I+zK) = \sum_{r=1}^{\infty} \frac{(-1)^{r+1}z^r}{r} \operatorname{Tr}_H(K^r).$$

The function $z\mapsto\det_{\!F}(I+zK)$ is entire, but its logarithm exists only locally away from its zeros after a branch has been chosen. Similarly, if $t\mapsto K(t)$ is differentiable in trace norm and $I+K(t)$ stays invertible, then along a chosen local branch

$$\frac{\mathrm d}{\mathrm dt} \log\det_{\!F}(I+K(t)) = \operatorname{Tr}_H \left[ (I+K(t))^{-1}K'(t) \right].$$

Writing $\operatorname{Tr}\log$ without these convergence, invertibility, and branch conditions hides precisely the questions this page is meant to answer.

A decision procedure

The input is a Hilbert space $H$, a candidate relative perturbation $K$, and estimates on its singular values. The main difficulty is proving summability; multiplying a large finite matrix is usually secondary.

Established fact about $K$	Licensed conclusion
$K\in\mathcal S_1$	$\operatorname{Tr}_H K$ and $\det_{\!F}(I+K)$ are canonical
$K=AB$ with $A,B\in\mathcal S_2$	$K\in\mathcal S_1$ by factorization
$K\in\mathcal S_2$ only	Hilbert–Schmidt control, but no ordinary trace or Fredholm determinant
$K$ compact only	$I+K$ is Fredholm, but no canonical scalar determinant follows
$K$ bounded only	No compactness, trace, or determinant conclusion

Use the following workflow.

1. Rewrite the proposed relative object as $I+K$. Do not start from a quotient of two unproved infinite determinants.
2. Prove $K\in\mathcal S_1$ from singular-value estimates, a diagonal model, or Hilbert–Schmidt factorization.
3. Define the trace or determinant only after that proof. Compute using finite rank, eigenvalues, the local trace–log series, or a derivative identity.
4. Approximate by $K_N\to K$ in trace norm and apply the continuity bound. Cross-check with an independent product, rank-one identity, or derivative.
5. Stop if only compactness, operator-norm convergence, or Hilbert–Schmidt control is available; name any added regularization rather than assigning the formal expression a value.

The output is a basis-independent trace or Fredholm determinant, an invertibility statement, and a convergence estimate tied to the topology that controls the quantity.

Where modified determinants begin

The words Fredholm operator and Fredholm determinant name different notions. If $K$ is compact, then $I+K$ is Fredholm and has an index. The ordinary scalar determinant used above needs the stronger condition $K\in\mathcal S_1$. A general Fredholm operator has no canonical scalar determinant merely because its kernel and cokernel are finite-dimensional.

For $K\in\mathcal S_2$, one can introduce the modified determinant

$$\det{}_2(I+K) = \det_{\!F}\!\left((I+K)e^{-K}\right),$$

because $(I+K)e^{-K}-I$ is trace class. If $K$ is also trace class, then

$$\det{}_2(I+K) = \det_{\!F}(I+K) \exp\!\left(-\operatorname{Tr}_H K\right).$$

This construction cancels the linear term in the formal trace–log expansion, even when that term is not itself traceable, and its multiplication law contains correction factors. It is therefore a regularized determinant, not evidence that the ordinary determinant existed all along. Higher modified determinants, zeta determinants, heat-kernel subtractions, determinant lines, and renormalized QFT determinants belong to later treatments (Kostenko 2019, §3.6, PDF; Simon 2005, Chapter 9).

If $\det_{\!F}(I+K)=0$, a logarithm is unavailable. Removing zero modes to form a “primed determinant” is another declared construction, not an algebraic cancellation licensed here.

Controlled Euclidean determinant ratio in one dimension

Consider the positive Dirichlet operator

$$H_0=-\frac{\mathrm d^2}{\mathrm d\tau^2}$$

on $L^2(0,L)$ with domain $H^2(0,L)\cap H_0^1(0,L)$. Its normalized eigenfunctions and eigenvalues are

$$e_n(\tau) = \sqrt{\frac2L}\sin\!\left(\frac{n\pi\tau}{L}\right), \qquad H_0e_n = \left(\frac{n\pi}{L}\right)^2e_n.$$

For $m\geq0$, let $H_m=H_0+m^2I$. Neither $\det H_0$ nor $\det H_m$ is being defined. Instead, the bounded relative operator is

$$H_mH_0^{-1} = I+K, \qquad K=m^2H_0^{-1}.$$

Here $H_0^{-1}$ is bounded and maps $L^2(0,L)$ into $\mathcal D(H_0)$, so the displayed composition is everywhere defined and bounded.

Its eigenvalues are

$$\kappa_n = \left(\frac{mL}{n\pi}\right)^2,$$

and

$$\|K\|_1 = \sum_{n=1}^{\infty}\kappa_n = \frac{(mL)^2}{6}.$$

Thus $K$ is trace class and the relative determinant

$$\det_{\mathrm{rel}}(H_m,H_0) := \det_{\!F}(H_mH_0^{-1})$$

exists without a zeta or heat-kernel prescription. Euler’s product

$$\frac{\sinh x}{x} = \prod_{n=1}^{\infty} \left(1+\frac{x^2}{n^2\pi^2}\right)$$

provides the independent exact evaluation

$$\det_{\mathrm{rel}}(H_m,H_0) = \prod_{n=1}^{\infty} \frac{(n\pi/L)^2+m^2}{(n\pi/L)^2} = \frac{\sinh(mL)}{mL},$$

where the value at $m=0$ is understood by continuity.

This is the trace-class interpretation of the Dirichlet ratio independently evaluated in Dunne 2008, §3, Eqs. (12)–(13).

Let $P_N$ retain the first $N$ Dirichlet modes. The normalized finite-dimensional Euclidean Gaussian ratio for one real bosonic variable per mode is

$$\frac{Z_{m,N}}{Z_{0,N}} = \left[ \frac{\det(P_NH_mP_N)} {\det(P_NH_0P_N)} \right]^{-1/2} = \prod_{n=1}^{N} \left(1+\kappa_n\right)^{-1/2}.$$

Because $P_NKP_N\to K$ in trace norm, these ratios converge to the positive branch

$$\frac{Z_m}{Z_0} = \det_{\mathrm{rel}}(H_m,H_0)^{-1/2}.$$

Accordingly, the relative one-loop Euclidean Gaussian contribution is

$$\Delta\Gamma_E^{(1)} = \frac12 \log\det_{\mathrm{rel}}(H_m,H_0) = \frac12 \log\!\left(\frac{\sinh(mL)}{mL}\right).$$

This is the controlled algebraic content of the Gaussian ratio, not a construction of a general continuum path-integral measure.

A derivative gives another check. Along $m>0$,

$$\begin{aligned} \frac{\partial}{\partial m^2} \log\det_{\mathrm{rel}}(H_m,H_0) &= \operatorname{Tr}_H(H_m^{-1})\\ &= \frac{L}{2m}\coth(mL) -\frac{1}{2m^2}, \end{aligned}$$

which agrees with differentiating the closed form.

The dimensional stop test

On a $d$-dimensional bounded box with Dirichlet boundary conditions, take a nonzero mass shift $m>0$. The same perturbation has relative part $K=m^2H_0^{-1}$. Its large-mode singular values scale as $|\mathbf n|^{-2}$, so

$$K\in\mathcal S_p \quad\Longleftrightarrow\quad 2p>d.$$

The relative perturbation $K$ is therefore trace class only for $d<2$, so the ordinary Fredholm determinant is available only in $d=1$. In $d=2,3$, $K$ is Hilbert–Schmidt but not trace class; in $d\geq4$, it is not even Hilbert–Schmidt, although the finite-volume inverse remains compact. The finite cutoff determinants still exist, but the trace-norm convergence argument has failed.

This is an explicit stop rule, not a computational inconvenience. In the dimensions used by relativistic QFT, a formal determinant ratio generally requires a declared regularization and local counterterms. Finite volume does not by itself solve the ultraviolet problem, and infinite volume can also destroy compactness. The higher-dimensional divergent product and its renormalized replacement are exhibited in Dunne 2008, §4, Eqs. (26)–(29).

Common pitfalls

A convergent-looking diagonal is called a trace. Basis independence, not one favorable ordering, is the issue. Prove trace class before using an infinite diagonal sum.

Every compact perturbation is assigned a determinant. Compactness makes $I+K$ Fredholm, but the ordinary Fredholm determinant requires $K\in\mathcal S_1$.

Singular values are multiplied in the determinant. Singular values test summability. The Fredholm product uses the generally complex eigenvalues, with algebraic multiplicity.

A formal kernel diagonal is integrated. An $L^2$ kernel has no canonical pointwise diagonal. A diagonal trace formula needs additional hypotheses.

Finite matrices are assumed to converge in the wrong norm. Strong, operator-norm, or Hilbert–Schmidt convergence does not control ordinary determinants. The $D_{1/n}$ example makes the failure explicit.

A global equality $\log\det=\operatorname{Tr}\log$ is used. A logarithm needs an invertible path and a branch; the power series also needs a convergence condition.

A modified determinant is treated as the ordinary one. The subtraction in $\det_2$ changes the object and its algebra. State the prescription and its correction terms.

Exercises

Schatten classification. Let $D_\alpha e_n=n^{-\alpha}e_n$ with $\alpha>0$. For which $p$ is $D_\alpha\in\mathcal S_p$?

Solution

The singular values are $n^{-\alpha}$, so

$$\|D_\alpha\|_p^p = \sum_{n=1}^{\infty}n^{-\alpha p}.$$

The $p$-series converges exactly when $\alpha p>1$. In particular, $D_\alpha$ is trace class exactly when $\alpha>1$ and Hilbert–Schmidt exactly when $\alpha>1/2$.

Rank-one trace and determinant. For $R=|u\rangle\langle v|$, compute $\operatorname{Tr}_H R$ and $\det_{\!F}(I+R)$.

Solution

Every rank-one operator is trace class. In an orthonormal basis, completeness gives

$$\begin{aligned} \operatorname{Tr}_H R &= \sum_n \langle e_n|u\rangle \langle v|e_n\rangle\\ &= \langle v|u\rangle. \end{aligned}$$

The only possible nonzero eigenvalue is $\langle v|u\rangle$, counted once when nonzero. Hence

$$\det_{\!F}(I+R)=1+\langle v|u\rangle.$$

The formula also holds when the eigenvalue vanishes, including a nonzero nilpotent rank-one operator.

Dimensional stop check. For $m>0$ and $K=m^2(-\Delta_D)^{-1}$ on a bounded $d$-dimensional box, decide when the ordinary determinant, only a Hilbert–Schmidt modified determinant, or neither test is available.

Solution

The singular values scale as $|\mathbf n|^{-2}$. Lattice counting gives $K\in\mathcal S_p$ exactly when $2p>d$. Thus $K\in\mathcal S_1$ only for $d=1$, so the ordinary Fredholm determinant is licensed only there. For $d=2,3$, $K\in\mathcal S_2\setminus\mathcal S_1$, so an explicitly modified $\det_2$ can be introduced but the ordinary determinant cannot. For $d\geq4$, this $\mathcal S_2$ test also fails; a different regularized construction would require additional analysis.

From determinants to QFT applications

The result established here is the operator-class criterion behind a determinant ratio: reduce it to $I+K$, prove $K\in\mathcal S_1$, and validate the Fredholm determinant in trace norm. Continue to Integrating Out Heavy Fields for the physical interpretation of determinant factors generated by integrating out fields. Burgess 2020, §2.1.2, p. 24, and §2.3.1, p. 33 is a concise reference for that one-loop and heavy-mode interpretation.

That destination develops bosonic and fermionic powers and phases, gauge fixing and ghosts, zero and negative modes, Lorentzian contours, scale separation, the large-mass or derivative expansion, matching, and local counterterms. This page does not imply that a formal QFT $\operatorname{Tr}\log$ is an ordinary trace or that a Fredholm determinant removes the need for renormalization.

References

• Folkmar Bornemann, “On the Numerical Evaluation of Fredholm Determinants”, Mathematics of Computation 79 (2010), §§2–5, especially Eqs. (2.3)–(2.8), (3.1)–(3.5), (4.1), and Theorem 5.1. This is the independent comparison for kernel cautions, equivalent determinant definitions, trace-norm perturbation bounds, and convergence of finite-rank projections.
• C. P. Burgess, Introduction to Effective Field Theory, Cambridge University Press, 2020, §2.1.2 printed p. 24 and §2.3.1 printed p. 33. This is the physical-handoff source for one-loop determinant terms and integrating out heavy modes. Its Lorentzian $+(i/2)\log\det$ convention is not used in the Euclidean calculation above.
• Gerald V. Dunne, “Functional Determinants in Quantum Field Theory”, Journal of Physics A 41 (2008) 304006, §1, §3 pp. 4–5 and Eqs. (9)–(13), §4 p. 7 and Eqs. (26)–(29), and §6 p. 9. These sections derive the Gaussian determinant, the one-dimensional Dirichlet ratio, the higher-dimensional divergence, and the zero-mode boundary. Heat-kernel and zeta constructions discussed there are not imported as ordinary Fredholm determinants.
• Aleksey Kostenko, Trace Ideals with Applications, PDF, lecture notes for the 2018–2019 advanced courses, Chapter 3: §§3.1–3.3, printed pp. 21–34; §§3.4.3–3.5, printed pp. 38–47; and §3.6, printed pp. 48–50. These sections develop singular values, trace and Hilbert–Schmidt classes, Fredholm determinants, trace-norm differentiation, and the regularized boundary. The notation has been translated to $\mathcal S_p$, $\det_{\!F}$, and the site’s conjugate-linear-bra convention.
• Barry Simon, Trace Ideals and Their Applications, second edition, Mathematical Surveys and Monographs 120, American Mathematical Society, 2005, Chapters 2–3, 5, and 9 (printed pp. 17–36, 45–52, and 75–80). These chapters establish Schatten ideals, Lidskii’s theorem, the ordinary determinant, Fredholm theory, and the distinction from regularized determinants.