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This is the definition of a tensor field I'm familiar with:

A tensor field $T$ of type $(k,l)$ is is a $C^{\infty}(M)$-multilinear map $T: \Omega^1(M) \times ...\Omega^1(M) \times \mathfrak{X}(M) \times... \times\mathfrak{X}(M)\rightarrow C^{\infty}(M)$. It eats $k$ covector fields and $l$ vector fields and spits out a smooth real function defined on a manifold $M$.

Now I came across the following claim:

The torsion $T$ of an affine connection $\nabla$ is a (1,2) tensor field. To check this it's sufficient to show that that the map defined by $T$ is $C^{\infty}(M)$-bilinear.

However, this doesn't make sense to me. A torsion tensor takes two vector fields so why is it a (1,2) tensor field? Where is the third argument? Also to check if $T$ is indeed a (1,2) tensor field, shouldn't we show that it is trilinear?

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    $\begingroup$ It eats two vectors and spits out one vector. $\endgroup$ – Deane Jan 22 at 1:01
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    $\begingroup$ In general, a tensor in $W\otimes V_1^*\otimes\cdots\otimes V_q^*$ can be thought of as a multilinear map $V_1\times\cdots\times V_q \rightarrow W$. $\endgroup$ – Deane Jan 22 at 1:24
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Associated to $T : \mathfrak{X}(M)\times\mathfrak{X}(M) \to \mathfrak{X}(M)$ is a map $T' : \Omega^1(M)\times\mathfrak{X}(M)\times\mathfrak{X}(M) \to C^{\infty}(M)$ given by

$$T'(\alpha, X, Y) = \alpha(T(X, Y)).$$

When people say that $T$ is a $(1, 2)$-tensor field, they really mean the associated map $T'$ is a $(1, 2)$-tensor field. To check that $T'$ is a tensor field, you need to check $C^{\infty}(M)$-linearity in all three arguments, but note that it is linear in the first argument as

\begin{align*} T'(f\alpha + g\beta, X, Y) &= (f\alpha + g\beta)(T(X, Y))\\ &= f\alpha(T(X, Y)) + g\beta(T(X, Y))\\ &= fT'(\alpha, X, Y) + gT'(\beta, X, Y). \end{align*}

So we only need to check that $T'$ is $C^{\infty}(M)$-linear in the second and third arguments, but this is the case if and only if $T$ is $C^{\infty}(M)$-linear in its two arguments.

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  • $\begingroup$ Ok, so $\alpha$ in this case is an arbitrary covector field on $M$? $\endgroup$ – mathripper Jan 22 at 10:45
  • $\begingroup$ @mathripper: Yes, also known as a differential one-form. $\endgroup$ – Michael Albanese Jan 22 at 11:40

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