4
$\begingroup$

If I have a $(p,q)$ tensor field that acts on $p$ covector fields and $q$ vector fields then does $T\left(X_1,\dots,X_p,Y_1,\dots,Y_q\right)$ return a function $f$ defined on the manifold by $$f\left(x\right)=T_x\left(X_1\left(x\right),\dots,X_p\left(x\right),Y_1\left(x\right),\dots,Y_q\left(x\right)\right)?$$ $T_x$ is now a $\left(p,q\right)$ tensor and $X_i\left(x\right)$ and $Y_j\left(x\right)$ are now covectors and tangent vectors respectively so this seems to make sense.

$\endgroup$
1
  • $\begingroup$ Holy cow, I had this exact same question dude $\endgroup$ May 4, 2021 at 11:31

1 Answer 1

4
$\begingroup$

That is correct. If $T$ is a smooth $(p, q)$-tensor field on $M$, then $T$ is a multilinear map

$$T : \underbrace{\Omega^1(M)\times\dots\times\Omega^1(M)}_{p\ \text{copies}} \times\underbrace{\mathfrak{X}(M)\times\dots\times\mathfrak{X}(M)}_{q\ \text{copies}} \to C^{\infty}(M)$$

where $\Omega^1(M)$ is the collection of covector fields (i.e. one-forms) on $M$, and $\mathfrak{X}(M)$ is the collection of vector fields on $M$. This approach to tensor fields is taken in Lee's Riemannian Manifolds: An Introduction to Curvature for example.

$\endgroup$

You must log in to answer this question.

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