# Bitensors

Bitensors aren't a very common feature of differential geometry, but they occasionally show up in various applications of relativity, to define distance functions, parallel transport, heat kernels, propagators and so on. Overall, they are very rarely described in the common bundle language of differential geometry, and never in any details. Here is an attempt to root them in a bit more firm ground.

A bitensor is, as the name implies, a tensor stemming from two different spacetime points. If we have the bitensor field $B$, it is a map

\begin{equation} B: M \times M \to V \end{equation}For $V$ the vector space we're considering as our tensor space. While they are mostly written in the obvious fashion, as a tensor field of two variables $T^{abc\ldots}(x,y)$, it should be kept in mind that they aren't quite so trivial as this could indicate as bundles, in particular because the two variables mean some caution should be used regarding derivatives of that field : the derivative can be performed on either point.

As far as rigorous definitions of the bitensor bundle go, they all agree on one thing : a bitensor bundle is an external tensor product bundle. So let's see in some more details what that is. As a roadmap, the definition of the external tensor product is that, for two vector bundles $\mathcal{V}_1$ and $\mathcal{V}_2$,

\begin{equation} \mathcal{V}_1 \boxtimes \mathcal{V}_2 = \mathrm{pr}_1^* \mathcal{V}_1 \otimes \mathrm{pr}_2^* \mathcal{V}_2 \end{equation}which is a product bundle of two pullback bundles. Let's see how to construct these first.

As a note first, due to the heavy use of products and bundles, and an attempt to keep everything somewhat rigorous, there will be a *lot* of projection operators, so to keep track :

- $\pi_{V}$ is the projection of a fiber bundle $\mathcal{V}$ with typical fiber $V$.
- $p_i$ is the projection of a product of two spaces $A_1 \otimes A_2$ to the $i$th space.

## 1. Products of vector bundles

Before we get too deep into the bitensor bundle itself, let's consider the usual tensor product of two vector bundles. Let's consider two vector bundles,

\begin{equation} \pi_{V_i} : \mathcal{V}_i \to M \end{equation}with $i = 1, 2$. We define the tensor product of those two bundles as

\begin{equation} \mathcal{V}_1 \otimes \mathcal{V}_2 = \bigcup_{p \in M} E_{1q} \otimes E_{2q} \end{equation}with the projection map

\begin{eqnarray} \pi_{V_1 \otimes V_2} : \mathcal{V}_1 \otimes \mathcal{V}_2 &\to& M\\ (v_1, v_2) &\mapsto& q \end{eqnarray}such that $\pi_{V_i} (v_i) = q$. The fiber at a point is then simply the tensor product of the two vector spaces. If we have the vector bundle atlases $\{ U_{\alpha}, \psi_{1, \alpha} \}$ and $\{ U_{\alpha}, \psi_{2, \alpha} \}$ on $\mathcal{V}_1$ and $\mathcal{V}_2$, we define

\begin{eqnarray} \Psi_{1, \alpha} \otimes \Psi_{2, \alpha} : (\mathcal{V}_1 \otimes \mathcal{V}_2)|_{U_\alpha} &\to& U_{\alpha} \times (V_1 \times V_2)\\ e &\mapsto& (\pi(e), (\Phi_{1, \alpha} \otimes \Phi_{2, \alpha})(e)) \end{eqnarray}...

Given any two sections of our two bundles $s_i \in \Gamma(\mathcal{V}_i)$, we can extend these to a section on the tensor product bundle via

## 2. Tangent bundles of products

On the flip side, let's now consider the tangent bundle (and more generally the tensor bundle) of the product of two manifolds. Let's consider two manifolds $M_1$, $M_2$, and their product, $M_1 \times M_2$. we define the projection functions,

\begin{eqnarray} \mathrm{pr}_i : M_1 \times M_2 &\to& M_i\\ (q_1, q_2) &\mapsto& q_i \end{eqnarray}Those projection maps, along with the universal property[5], define the product of the two manifolds. The universal property here being that for every object $X$ and every pair of maps $f_1 : X \to M_1$, $f_2 : X \to M_2$, there exists a unique function $f : X \to M_1 \times M_2$, such that $\mathrm{pr}_i \circ f = f_i$.

We also define two families of insertion maps,

\begin{eqnarray} \iota_{q_1} : M_2 &\to& M_1 \times M_2\\ q_2 &\mapsto& (q_1, q_2)\\ \iota'_{q_2} : M_1 &\to& M_1 \times M_2\\ q_1 &\mapsto& (q_1, q_2) \end{eqnarray}Those functions obey the relations

\begin{eqnarray} \mathrm{pr}_1 \circ \iota'_{q_2} &=& \mathrm{Id}_{M_1}\\ \mathrm{pr}_2 \circ \iota_{q_1} &=& \mathrm{Id}_{M_2}\\ \mathrm{pr}_1 \circ \iota_{q_1} &=& q_1 \\ \mathrm{pr}_2 \circ \iota'_{q_2} &=& q_2 \end{eqnarray}The first two being automorphisms on $M_1$ and $M_2$, the last two mapping $M_2$ to $M_1$ and vice-versa.

Let's now define some pushforwards of our functions. From differential geometry, we know that we can define the following pushforwards between the various tangent bundles :

\begin{eqnarray} \mathrm{pr}_{i*} : T(M_1 \times M_2) &\to& TM_i \end{eqnarray}By the chain rule of the pushforward, we have that, given two functions $f : L \to M$ and $g : M \to N$, with the pushforward $f_{*p} : T_p L \to T_{f(p)} M$ and $g_{*q} : T_q M \to T_{g(q)} N$, the pushforward composition $(g \circ f)_{p} : T_p L \to T_{g(f(p))} N$ is given by

\begin{eqnarray} (g \circ f)_{*p} = g_{*f(p)} \circ f_{*p} \end{eqnarray}We have two kinds of maps to consider here, the identity automorphism and a constant map to a given point. Using the usual definition of the pushforward, we simply get, for some vector field $X$ and function $f$,

\begin{eqnarray} (\mathrm{Id})_{*p}(X)[f] &=& X[f \circ \mathrm{Id}]\\ &=& X[f]\\ (c)_{*p}(X)[f] &=& X[f \circ c]\\ &=& X[f(q)]\\ &=& 0 \end{eqnarray}So that the pushforward of the identity on the manifold is the identity on the tangent bundle, and the pushforward of a constant function simply maps to the zero vector. Given all this, we get

\begin{eqnarray} \mathrm{pr}_{1*(q_1, q_2)} \circ \iota'_{q_2*q_1} &=& \mathrm{Id}_{T_{q_1}M_1}\\ \mathrm{pr}_{2*(q_1, q_2)} \circ \iota_{q_1*q_2} &=& \mathrm{Id}_{T_{q_2}M_2}\\ \mathrm{pr}_{1*(q_1, q_2)} \circ \iota_{q_1*q_2} &=& 0\\ \mathrm{pr}_{2*(q_1, q_2)} \circ \iota'_{q_2*q_1} &=& 0 \end{eqnarray}This is 4 out of 5 of the definition of a direct sum of modules, only missing

\begin{eqnarray} \iota'_{q_2*q_1} \circ p_{1*(q_1, q_2)} + \iota_{q_1*q_2} \circ p_{2*(q_1, q_2)} = \mathrm{Id}_{T_(q_1, q_2)(M_1 \times M_2)} \end{eqnarray}Let's consider the map

\begin{eqnarray} \iota'_{q_2*q_1} + \iota_{q_1*q_2} : T_{q_1} M_1 \times T_{q_2} M_2 &\to& T_{(q_1, q_2)}(M_1 \times M_2) \end{eqnarray}...

The tangent spaces of our product of manifolds at a point is therefore the direct sum of the tangent spaces at those equivalent points. With a bit more work, if we also had the maps $f_i : X \to M_i$ and $g_i : M_i \to Y$, we can show the full diagram of the universal property of direct sums,

\begin{equation} \array{ &T_xX&\\ \overset{f_{1*}}{\swarrow}&\downarrow_{f_1 \oplus f_2}&\overset{f_{2*}}{\searrow}\\ T_{q_1} M_1 & \underset{\mathrm{pr}_{1*}}{\overset{\iota'_{q_2*}}{\rightleftarrows}} T_{q_1} M_1 \oplus T_{q_2} M_2 \underset{\mathrm{pr}_{2*}}{\overset{\iota_{q_1*}}{\leftrightarrows}} & T_{q_2} M_2 \\ \underset{g_{1*}}{\searrow} & \downarrow_{ g_{1*} \oplus g_{2*}} & \underset{g_{2*}}{\swarrow} \\ &T_yY&} \end{equation}## 3. The external tensor product bundle

External tensor product bundles do have some literature, but finding some geared towards physics is a bit challenging, so let's go on describing them without using too much category theory.

Let's consider two vector bundles, $\pi_1 : \mathcal{V}_1 \to M_1$ and $\pi_2 : \mathcal{V}_2 \to M_2$, each of these with the typical fiber $V_i = \pi_i^{-1}(\{ q_i \})$. First, let's consider the space $M_1 \times M_2$, the product manifold of our two base spaces. We define on this two projectors, $\mathrm{pr}_1$ and $\mathrm{pr}_2$, such that

\begin{eqnarray} \mathrm{pr}_i : M_1 \times M_2 &\to& M_i\\ (q_1, q_2) &\mapsto& q_i \end{eqnarray}Via each of these maps, we can define a pullback bundle, $\mathrm{pr}_i^* \mathcal{V}_i$

\begin{equation} \mathrm{pr}_i^* \mathcal{V}_i = \{ ( (q_1, q_2), e_i ) \in (M_1 \times M_2) \times \mathcal{V}_i | \mathrm{pr}_i(q_1, q_2) = \pi_i(e_i) \} \end{equation}with projection

\begin{equation} \pi_{\mathrm{pr}_i^* \mathcal{V}_i} ( (q_1, q_2), e_i ) = (q_1,q_2) \end{equation}The fiber at $(q_1, q_2)$ is the set of values $e_i$ such that $\pi_i(e_i) = q_i$. In other words, it is isomorphic to $V_i$.

One thing we will later need is the pushforward of the projection. Given a vector field $X$ on $T(M_1 \times M_2)$, we have

\begin{equation} (\mathrm{pr}_{i*} X)_{(q_1,q_2)}[f] = X[\mathrm{pr}_i^* f] = X[f(\mathrm{pr}_i(q_1,q_2))] \end{equation}If we have the local basis $\partial_{\mu^i}$ of $TM_i$, for coordinates $x^i$ of $M_i$, then our pushforward is

\begin{equation} (\mathrm{pr}_{i*} X)_{(q_1,q_2)}[f] = X^{\mu^i}(q_1,q_2) [\partial_{\mu^i}f( \mathrm{pr}_i(x^1, x^2))] \end{equation}in other words, we only perform the derivative in one coordinate. Our pushforward is such that

Our bitensor bundle is the tensor product of our two pullback bundles :

\begin{equation} \mathcal{V}_1 \boxtimes \mathcal{V_2} = \mathrm{pr}_1^* \mathcal{V}_1 \otimes \mathrm{pr}_2^* \mathcal{V}_2 \end{equation}## Bitensors

With all this, our bitensors are simply sections of the exterior product of tensor bundles. If we have two tensor bundles ${T^{r}}_{s}M$ and ${T^{r'}}_{s'}M$ over the same base space, then a bitensor $B$ belongs to

\begin{equation} B \in \Gamma({T^{r}}_{s}M \boxtimes {T^{r'}}_{s'}M) \end{equation}The simplest case is the biscalar field, which is simply the product of two $(0,0)$ tensor fields, in other words the line bundle.

\begin{equation} B \in \Gamma(LM \boxtimes LM) \end{equation}The tensor product of two copies of the canonical line bundles being simply isomorphic to the canonical line bundle.

## Covariant derivative of bitensors

Taking the usual covariant derivative on the tangent bundle, the covariant derivative of bitensors is simple enough to derive. First, let's consider the covariant derivative on a bundle product. For a section of $\mathcal{V}_1 \otimes \mathcal{V}_2$ with a section of the form $v_1 \otimes v_2$, its canonical covariant derivative, $\nabla : \mathcal{V}_1 \otimes \mathcal{V}_2 \to \Omega^1(M) \otimes \mathcal{V}_1 \otimes \mathcal{V}_2$, is obtained by the Leibniz rule :

\begin{equation} \nabla(v_1 \otimes v_2) = (\nabla v_1) \otimes v_2 + v_1 \otimes (\nabla v_2) \end{equation}Or, using our projectors,

\begin{equation} \nabla(v) = (\nabla \text{pr}_1(v)) \otimes \text{pr}_2(v) + \text{pr}_1(v) \otimes (\nabla \text{pr}_2(v)) \end{equation}Therefore we simply need to define the action of our derivative on the pullback bundle. Given a connection $\nabla$ on $E$, and a pullback map $\phi : $, the pullback connection $\phi^* \nabla$ on $\phi^* E$ is defined by

\begin{equation} (\phi^* \nabla)_X (\phi^* s) = \phi^* (\nabla_{d\phi(X)} s) \end{equation}In our case, the connection will be a tensor connection over $M_i$

7In most cases when dealing with bitensors, we will be interested in the derivative of only one variable. As for a

[Prove the commutativity of the 1 and 2 derivatives]## Bibliography

Last updated :

*2022-05-23 13:12:57*