I'm interested in finding the appropriate geometric construct for the integration of symmetric tensors, analogous to the way differential forms can be integrated over manifolds.

The motivation comes from gauge theories, commonly used in theoretical physics. The simplest example is a vector field $A_i$ defined on a Euclidean space $R^m$, where the vector field is only defined up to addition of a gradient, $A_i \sim A_i + \partial_i \alpha$, for some scalar $\alpha$. The meaningful quantities are the integrals of $A_i$ around closed curves, which eliminates any contribution from the gradient term. More generally, allow $A$ to be an $n$-form defined on $R^m$ ($m>n$), defined only up to addition of an exact form, $A\sim A+d\alpha$. The meaningful quantities are the integrals of $A$ over closed $n$-dimensional submanifolds of $R^m$, which kill the contribution from the exact form.

Now, to the main question. Allow the gauge field to be a symmetric tensor $A_{ij}$ defined on the Euclidean space, but only up to a second derivative, $A_{ij}\sim A_{ij}+\partial_i\partial_j \alpha$ for arbitrary scalar $\alpha$. I'm wondering what the appropriate geometric construct is for integrating symmetric tensors which would allow me to obtain meaningful quantities, independent of $\alpha$.

I should mention that I have indeed been able to find appropriate "objects" of integration when the theory is defined on a lattice approximation to $R^m$, so I believe the concept to be well-defined. I just need a better way to mathematically characterize these objects in order to make meaningful statements in the continuum limit. Based on the lattice analysis, it seems that the "objects" of integration have well-defined dimension (based on notions of self-similarity, as in fractals), but do not appear to be conventional manifolds. Furthermore, the "dimension" of the objects can actually exceed $m$ in certain cases, leading to a picture in which higher-dimensional objects have been embedded in a lower-dimensional Euclidean space. But these are merely conjectures at this point.

Does anyone have any ideas as to what mathematical construct I am looking for?

  • 1
    $\begingroup$ The "second derivative" is not well defined until you specify a connection. (It is not tensorial/coordinate invariant). I am not sure how you want to think of $A$: is it an object subordinate to the chosen connection on $T^*R^{m}$. $\endgroup$ Jul 17, 2016 at 2:11
  • $\begingroup$ Why can't you just form $I_i=\int_C A_{ij} dx^j$? $\endgroup$ Jul 17, 2016 at 2:28
  • $\begingroup$ Thanks for the comments. To Willie, in general I know derivatives should have connection terms to be tensorial. For the time being, I'm only interested in the trivial connection. But I would also be interested to know the answer in the more general case. Simply replace the second derivative in the question above with the appropriate covariant derivatives. $\endgroup$
    – M. Pretko
    Jul 17, 2016 at 5:08
  • $\begingroup$ To Peter, the integral you mentioned definitely exists as a valid subcase. My analysis so far indicates that this is just one degenerate subcase of a wider class of objects. I want to find a general scheme of integration which takes a symmetric tensor as input and then outputs a scalar quantity. Of course one can simply contract one index of the tensor with an arbitrary vector, $\tilde{A}_i = v_j A^{ij}$, reducing the problem to the vector case. But this does not appear to be the most general solution. $\endgroup$
    – M. Pretko
    Jul 17, 2016 at 5:17

1 Answer 1


Here is one way to construct all "local" gauge-invariant quantities out of a symmetric tensor $A_{ij}$. It would be up to you to decide how it meshes with the intuition you gained from your investigations on the lattice.

First, define $R[A]_{ij:kl} = \partial_i\partial_k A_{jl} - \partial_j\partial_k A_{il} - \partial_i\partial_l A_{jk} + \partial_j\partial_l A_{ik}$ and $C[A]_{ij:k} = \partial_i A_{jk} - \partial_j A_{ik}$. You might notice that $R[A]$ is the linearized Riemann tensor, applied to a perturbation $\eta_{ij} \mapsto \eta_{ij} + A_{ij}$ of the flat Euclidean metric (or of whatever signature you like), while $C[A]$ is not exactly the linearized Christoffel connection, but a certain projection of it.

It is a straight forward exercise to show that a pure gauge field configuration $A_{ij} = \partial_i \partial_j \alpha$ is annihilated by both $R[A]$ and $C[A]$. But it is also true that any $A_{ij}$ annihilated by both $R[A]$ and $C[A]$ must be of that form, at least locally. First, $R[A] = 0$ implies that $A_{ij} = \partial_i v_j + \partial_j v_i$ for some $v_i$ (at least locally), which is essentially an infinitesimal restatement of the well-known fact that any metric that has zero Riemann curvature admits coordinates in which it takes the standard Euclidean form. Next, for $A_{jk} = \partial_j v_k + \partial_k v_j$ we have $C[A]_{ij:k} = \partial_k (\partial_i v_j - \partial_j v_i)$. By applying the Poincaré lemma, $C[A] = 0$ means that (locally) $v_i = \partial_i \alpha + B_{ij} x^j$, where $B_{ij} = B_{[ij]}$ is constant and the $x^j$ are Euclidean coordinates. But that last term is irrelevant, since we have found that our original $A_{ij} = \partial_i v_j + \partial_j v_i = \partial_i \partial_j (2\alpha)$ (at least locally).

Now consider any integral of the form \begin{equation} F(A) = \int ( f^{ijkl}(x) R[A]_{ij:kl} + g^{ijk}(x) C[A]_{ij:k} ) \, dx . \end{equation} For example, the tensors $f$ and $g$ could be arbitrary test functions (smooth and compactly supported), but could also be distributions supported on lower dimensional submanifolds, whatever makes sense in your context. The point is that all (reasonable) linear gauge-invariant functionals of the $A_{ij}$ field can be written in the above form.

  • $\begingroup$ Hi Igor. Thanks for the response! This is helpful to me. But I have one question here. In the integral $F(A)$ you've written, taking $A_{ij} = \partial_i\partial_j \alpha$ will make the integrand itself vanish. This is analogous to, in the vector case, looking at $\int dx_i \epsilon^{ijk}\partial_j A_k$. What I'm wondering is if there is also an analog of $\int dx_i A^i$, where the integrand does not vanish locally upon taking a "pure gauge" potential, but only vanishes upon integration. $\endgroup$
    – M. Pretko
    Jul 17, 2016 at 16:52
  • $\begingroup$ In the usual case of the vector potential, you can link the two kinds of integrals using Sokes' theorem: $\int_{\partial S} dx^i A_i = \int_S dx^i\wedge dx^j \partial_{[i} A_{j]}$. The fact that you have 1- and 2-dimensional integrals is not important here, because you can always write them as volume integrals against an appropriately chosen distribution. Analogously, you can use integration by parts to rewrite $F(A) = \int (R^*[f]^{ij} + C^*[g]^{ij}) A_{ij}$ (or some an intermediate form of it), where $R^*$ and $C^*$ are the formal adjoint operators to $R$ and $C$. $\endgroup$ Jul 17, 2016 at 17:25
  • $\begingroup$ Hi Igor. Thanks for clarification. I understand you better now. The last point which bugs me is the need to specify tensors like $f$ and $g$ independently of the region of integration. In a normal gauge theory, the Wilson lines/surfaces already come equipped with the appropriate structures. For example, a vector $A_i$ can be contracted into the tangent vector $dx_i$ to the curve. Similarly, an antisymmetric $A_{ij}$ can be naturally integrated against the built in 2-form $dx^i \wedge dx^j$ defining a surface. $\endgroup$
    – M. Pretko
    Jul 17, 2016 at 22:26
  • $\begingroup$ The question here then becomes, do you know of any geometric object (not necessarily a manifold) which comes with a built-in symmetric tensor? This is what appears to be happening on the lattice. The Wilson "objects" are actually defined via a symmetric tensor at each location, analogous to a curve being specified by its tangent vector. The objects are weird-looking and tend to have lots of self-intersection, so I haven't been sure if manifold is the right concept. Any suggestions are welcome. Thanks again for the help! $\endgroup$
    – M. Pretko
    Jul 17, 2016 at 22:30
  • $\begingroup$ I can't guess at what you've found on the lattice, so there is no way I can make a comparison. I can only repeat any kind of integration against lower dimensional manifolds (or much more general objects like singular chains), whether defined "naturally" or not, is simply a special case of volume integrals against appropriately chosen distributions. Ex: for a curve $\gamma$, $\int_\gamma dx^i A_i(x) = \int dy \Gamma^i(y) A_i(y)$, with $\Gamma^i(x) = \int_\gamma dt \dot{\gamma}^i(t) \delta(y-\gamma(t))$. I'm sure you can imagine that this step could be reversed for appropriate $\Gamma^i(y)$. $\endgroup$ Jul 18, 2016 at 3:26

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.