A I know a restrictive $-$ though sufficiently general in physics $-$ definition of the Tensor object:

A $(p,q)-$Tensor is a multilinear function like: $$ T: V\times\cdot\cdot\cdot\times V\times V^{*}\times\cdot\cdot\cdot\times V^{*} \to \mathbb{K} \tag{1}$$

But a guy told me something astonishing, and something that I never encountered before. He said that we can define a tensor object such that the field $\mathbb{K}$ can be replaced by, for example, $\mathbb{K}^{n}$, or even $V^{*}$. In the latter case he said:

A $1-$Form Valued Tensor.

So I would like to know:

What exactly is an object like:

$$ T: V\times\cdot\cdot\cdot\times V\times V^{*}\times\cdot\cdot\cdot\times V^{*} \to V^{*} \tag{2}$$

They are tensors?


2 Answers 2


If $W$ is any vector space, you can talk about a $W$-valued $(p,q)$-tensor. By definition, it is just a multilinear map $T: V^{\times p} \times (V^*)^{\times q} \to W$. Cases of interest being $W = \Bbb K$, $W = V$ and $W = V^*$.

If you have a bilinear map $\mu\colon W_1 \times W_2 \to W_3$, where $W_i$ ($i=1,2,3$) are vector spaces, you can define the $\mu$-tensor product of a $W_1$-valued $(p,q)$-tensor and a $W_2$-valued $(p',q')$-tensor as a $W_3$-valued $(p+p',q+q')$-tensor via the formula $$(T\otimes_\mu S)(v_1,\ldots, v_{p+p'}, \xi^1, \ldots, \xi^{q+q'}) = \mu\left(T(v_1,\ldots, v_p, \xi^1,\ldots, \xi^q), S(v_{p+1},\ldots, v_{p+p'}, \xi^{q+1},\ldots, \xi^{q+q'})\right).$$The usual tensor product comes from the usual multiplication $\mu\colon \Bbb K \times \Bbb K \to \Bbb K$, $\mu(x,y)=xy$. This is particularly interesting when we focus on forms. More precisely, if $\alpha$ is a $W_1$-valued $k$-form and $\beta$ is a $W_2$-valued $\ell$-form, one obtains a $W_3$-valued $(k+\ell)$-form via $$\alpha \wedge_\mu \beta(v_1,\ldots, v_{k+\ell}) = \sum_{\sigma \in S_{k+\ell}} {\rm sgn}(\sigma) \mu\left(\alpha(v_{\sigma(1)},\ldots, v_{\sigma(k)}), \beta(v_{\sigma(k+1)},\ldots, v_{\sigma(k+\ell)})\right).$$It is not necessarily true that $\beta \wedge_\mu \alpha = (-1)^{k\ell} \alpha\wedge_\mu \beta$ anymore, as this sort of symmetry depends on $\mu$. Some very special cases are:

  1. $W_1 = \Bbb K$ and $W_3 = W_2$, with $\mu\colon \Bbb K \times W_2 \to W_2$ given by $\mu(\lambda, v) = \lambda v$. Here $\beta \wedge \alpha = (-1)^{k\ell} \alpha\wedge \beta$ is true.
  2. When $W_1 = W_2=W_3 = \mathfrak{g}$ is a Lie algebra and $\mu = [\cdot,\cdot]$ is the Lie bracket. Here we have the modified identity $\beta \wedge_{[\cdot,\cdot]} \alpha = (-1)^{k\ell+1} \alpha \wedge_{[\cdot,\cdot]} \beta$, as skew-symmetry of $[\cdot,\cdot]$ gives an extra minus sign.

On the manifold level, you can look at $E$-valued $(p,q)$-tensor fields on a manifold $M$, where $E \to M$ is any vector bundle over $M$. More precisely, such $E$-valued tensor fields are sections of the bundle $T^*M^{\otimes p}\otimes TM^{\otimes q}\otimes E$. In particular, for $1$-form valued tensors, $E = T^*M$.

If $E \to M$ is equipped with a linear connection $\nabla$ (even in the case $E = TM$, this need not necessarily be the Levi-Civita connection of any pseudo-Riemannian metric on $M$), one may mimic the classical Palais formula to define a covariant exterior derivative ${\rm d}^\nabla \colon \Omega^k(M;E) \to \Omega^{k+1}(M;E)$ via $$({\rm d}^\nabla \alpha)(X_0,\ldots, X_k) = \sum_{i=0}^k (-1)^i \nabla_{X_i}(\alpha(X_0,\ldots, \hat{X_i},\ldots, X_k)) + \sum_{i<j} (-1)^{i+j} \alpha([X_i,X_j], X_0,\ldots, \hat{X_i},\ldots, \hat{X_j},\ldots, X_k).$$This does not satisfy ${\rm d}^\nabla \circ {\rm d}^\nabla = 0$ --- this is true if and only if $\nabla$ is a flat connection. A section $\psi$ of $E \to M$ is the same thing as a $E$-valued $0$-form, and we have things like $({\rm d}^\nabla \psi)(X) = \nabla_X\psi$, $(({\rm d}^\nabla)^2\psi)(X,Y) = R^\nabla(X,Y)\psi$, and so on. If $E$ is a trivial vector bundle, then ${\rm d}^\nabla = {\rm d}$ is the usual exterior derivative.

When $E = TM$, the torsion $\tau^\nabla$ (given by $\tau^\nabla(X,Y) = \nabla_XY-\nabla_YX-[X,Y]$) can be seen as a $TM$-valued $2$-form, and a direct computation shows that $$({\rm d}^\nabla \tau^\nabla)(X,Y,Z) = R^\nabla(X,Y)Z + R^\nabla(Y,Z)X + R^\nabla(Z,X)Y,$$so we see that if $\nabla$ is torsion free, the first Bianchi identity holds, and its content is nothing more than "the derivative of zero equals zero" (in other words, the first Bianchi identity is an algebraic "accident" caused by lack of torsion).

Back to the general case $(E,\nabla) \to M$, one may see the curvature as a ${\rm End}(E)$-valued $2$-form. The second Bianchi identity can be phrased as ${\rm d}^\nabla R^{\nabla} = 0$ and is always true.

This sort of formalism is useful when studying gauge theories and connections in principal bundles, as a gauge field $A$ for a principal $G$-bundle $P \to M$ is nothing more than a certain $\mathfrak{g}$-valued $1$-form, and the gauge strength $F_A$ is a $\mathfrak{g}$-valued $2$-form defined from taking suitable derivatives of $A$.

This should give you enough context already, so I'll stop here.

  • 2
    $\begingroup$ This is a very nice answer (and I found your online notes on the exterior covariant derivatives very helpful when studying recently)! You probably already know this, but two other very nice instances of the bilinear pairing is composition $\circ:\text{End}(V)\times\text{End}(V)\to\text{End}(V)$ and evaluation $\epsilon:\text{End}(V)\times V \to V$. Then, the higher exterior covariant derivatives can be succinctly written for example as $(d^{\nabla})^{2p}\psi=(R\wedge_{\circ} \cdots \wedge_{\circ} R)\wedge_{\epsilon}\psi$ where we wedge the curvature $p$-times. $\endgroup$
    – peek-a-boo
    Jun 17, 2021 at 1:07
  • 2
    $\begingroup$ So, the the torsion identity can be written simply as $d^{\nabla}\tau = (d^{\nabla})^2(\text{id}_{TM})=R\wedge_{\epsilon}\text{id}_{TM}$. Then of course, if one wishes to extract a more explicit/usual formula, they can work through the definition of the wedge products as opposed to the Palais formula (which for some reason I find very Laborious). $\endgroup$
    – peek-a-boo
    Jun 17, 2021 at 1:08
  • $\begingroup$ I'm glad the notes were helpful. It never crossed my mind to use $\circ$ and $\epsilon$ as the pairings, but I like this, it's very elegant! $\endgroup$
    – Ivo Terek
    Jun 17, 2021 at 1:14

This is a standard, and some times very convenient, way to reinterpret tensors. Consider, for instance, a (1-1)-tensor $$ T:V\times V^*\to\Bbb K $$ Now take this tensor, and feed it an element of $V^*$. What do you have left? An object that takes an element from $V$ as input and gives you elements in $\Bbb K$ as outputs. These are exactly the elements of $V^*$. Which is to say, we can think of $T$ as a map $$ V^*\to V^* $$ rather than a map $V\times V^*\to\Bbb K$. Similarly, by using the fact that $(V^*)^*$ is naturally identified with $V$, we can also think of $T$ as a map $V\to V$.

Playing with different tensor valences and exactly what order tensors you feed them, you can get lots of different interpretations. The most common one is probably the metric tensor (or inner product, dot product, scalar product, whatever you want to call it), which is nominally $V\times V\to\Bbb K$. But it can similarly be reinterpreted as a map $V\to V^*$, and in this form it is known as the transpose (yes, mostly the same transpose that you know from linear algebra). If you are familiar with bras and kets in quantum mechanics, for instance, this is the map that transforms one into the other.


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .