I've read "We can only define the integral of a scalar function. The integral of a vector or tensor field is meaningless in curved spacetime" on many books and lectures on General Relativity (For example on this notes from NYU: https://cosmo.nyu.edu/yacine/teaching/GR_2018/lectures/integration.pdf) and I was wondering if there's a way around it. For example, let's say there is a geodesic $z^\mu(\tau)$ and we are integrating some bitensor $G_{\mu\nu}(z,z')$ to get some quantity $Q(x)$

\begin{equation} Q\left(x\right)=\int_{-\infty}^{\infty}G_{\mu\nu}\left[x,z\left(\tau'\right)\right]G^{\mu\nu}\left[x,z\left(\tau'\right)\right]d\tau' \end{equation}

Here we are integrating a scalar function (once the bitensor $G_{\mu\nu}$ gets contracted with $G^{\mu\nu}=g^{\mu\alpha}g^{\nu\beta}G_{\alpha\beta}$). Also, the resulting scalar $Q(x)$ is evaluated at a spacetime point $x^\mu$ which might or might not be on the geodesic $z^\mu(\tau)$. Now, we could also try to derive $Q(x)$

\begin{equation} \frac{\partial}{\partial x^{\rho}}Q\left(x\right)=\\=\frac{\partial}{\partial x^{\rho}}\int_{-\infty}^{\infty}G_{\mu\nu}\left[x,z\left(\tau'\right)\right]G^{\mu\nu}\left[x,z\left(\tau'\right)\right]d\tau'\end{equation}

This is also a well-defined object, we are integrating a scalar over the worldline and then deriving. Now, what happens if we take the derivative inside the integral?

\begin{equation} \frac{\partial}{\partial x^{\rho}}Q\left(x\right)=\int_{-\infty}^{\infty}\frac{\partial}{\partial x^{\rho}}\left\{ G_{\mu\nu}\left[x,z\left(\tau'\right)\right]G^{\mu\nu}\left[x,z\left(\tau'\right)\right]\right\} d\tau' \end{equation}

Now the right hand side is the integral of a tensor with one free index $\rho$. But we had said that $\frac{\partial }{\partial x^\rho}Q(x)$ is well defined. So is integrating over a tensor also ill-defined in this case? Or could we procede without thinking too much about it?

Note: The bitensor $G_{\mu\nu}$ here is not important, it's just an example to show what my question is. If you like think of it as a Green's function.

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    $\begingroup$ If your object depends on two spacetime points, then it is not a tensor, because (co)tangent spaces at different points are completely distinct objects with no natural isomorphisms between them. Instead you should use bitensors. $\endgroup$
    – A.V.S.
    Mar 30, 2021 at 5:30
  • 1
    $\begingroup$ Thanks for the comment! $G_{\mu\nu}$ is a bi tensor, actually. Let me edit that $\endgroup$ Mar 30, 2021 at 5:31
  • $\begingroup$ It seems v3 still depends on two spacetime points. $\endgroup$
    – Qmechanic
    Mar 30, 2021 at 16:14

2 Answers 2


In this example, you are not integrating a vector field over curved spacetime. You are integrating over a family of vectors that all "belong" to the same point in spacetime. That's why the integral works.

Let's take a step back, and first notice that we can simplify your (bi)tensor field down to a scalar field without changing the question: letting $G[x,x']=G_{\mu\nu}[x,x']G^{\mu\nu}[x,x'],Q[x]=\int_{-\infty}^\infty G[x,z^\mu(\tau)]\,d\tau,\partial_\mu=\frac{\partial}{\partial x^\mu},$ what is the meaning of $\partial_\mu Q[x]=\int_{-\infty}^\infty\partial_\mu G[x,z^\nu(\tau)]\,d\tau$ ? We should inspect what $\partial_\mu$ actually does. If you call our spacetime manifold $M$, the operator $\partial_\mu$ takes a scalar field (like our $Q:M\to\mathbb{R}$) and a point $x\in M$ and produces a vector $\partial_\mu Q[x]\in {T\!M}_x,$ where ${T\!M}_x$ denotes the tangent space of $M$ at $x.$ If you evaluate $\partial_\mu Q[p],\partial_\mu Q[q]$ at two different points $p\neq q\in M,$ then these vectors will live in different vector spaces ${T\!M}_p,{T\!M}_q,$ and there will not be a canonical way to add them up. That is why you've been told you can't integrate vector/tensor fields over curved spacetime.

But the thing is that your integral does not evaluate such a vector field at different points. On the LHS, we have $\partial_\mu Q$ evaluated at $x$, which produces a vector that lives in ${T\!M}_x$. On the RHS, we have $\partial_\mu G[x,z^\nu(\tau)].$ Where's the scalar field, and at which point is the derivative being evaluated? Well, since we're only taking the derivative with respect to $x$, the scalar field is actually $G[x,z^\nu(\tau)],$ considered only as a function of $x$ (certain mathematicians might denote it $G[-,z^\nu(\tau)]$, with a "gap"), and the derivative of this field is being evaluated only at $x$. Therefore, for all the $\tau$, the integrand in fact is a vector in ${T\!M}_x,$ and so the integral is well defined because you aren't actually integrating vectors "from" different points in curved spacetime. (Or, in other words, if you're considering how a field $G[x,z^\nu(\tau)]$ varies for changes in $x$ while holding $z^\nu(\tau)$ constant, the vector $\partial_\mu G[x,z^\nu(\tau)]$ representing that variation should be "anchored" at only $x,$ and if you vary $\tau$ you can get a family of such vectors which you can then add up.)


What you are forgetting is if one asks about integration it means summing over infinitesimal objects of the same kind. Like think about it as integrating a vector field in flat spacetime, the vectors belong to the same vector space like a scalar, as scalars belong to their groups because they are $(0,0)$ tensors, that is, they do not change under change of co-ordinates.

Consider curved space-time on the other hand. The scalars remain the same because of their properties. But what about a vector field in this space. The idea of Parallel transport and covariant deravitive tells that there cant be a constant vector field in this spacetime, often thought of manifolds. So how can you integrate in vectors belonging from different species. Same goes for tensor fields.

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    $\begingroup$ I understand why integrating tensors in curved space is problematic, in general. My question is if it's still problematic in the particular example I showed, where it is clear that what looks like integrating a tensor is just deriving the integral of a scalar function, which should be well defined. $\endgroup$ Mar 30, 2021 at 5:35
  • $\begingroup$ Do you remeber Christoffel symbols, that are made from deravitives and the inverse of a metric tensor (0, 2) while still it's not a tensor $\endgroup$ Mar 30, 2021 at 5:38
  • $\begingroup$ Maybe just maybe the same goes for this one as I am not well knowledgeable in terms of worldlines and bit tensors $\endgroup$ Mar 30, 2021 at 5:40

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