# Classifying tensor fields thought of as multilinear maps

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?

• It eats two vectors and spits out one vector. Commented Jan 22, 2021 at 1:01
• In general, a tensor in $W\otimes V_1^*\otimes\cdots\otimes V_q^*$ can be thought of as a multilinear map $V_1\times\cdots\times V_q \rightarrow W$. Commented Jan 22, 2021 at 1:24

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.
• Ok, so $\alpha$ in this case is an arbitrary covector field on $M$? Commented Jan 22, 2021 at 10:45