Tensor and Wedge

Question

I was wondering, what is the meaning of $$0\otimes dx \otimes dx \space\text{or}\space 0\wedge dx\wedge dy$$

These are some loose terms, and I have not defined any coordinate system, but just in general, what do they mean, and do they just equal to $0?$ Also what is the interior product of $0$ over some vector field $X$, $i_X(0)?$

EDIT: Let us say they are maps working with cotangent space and tangent space of the manifold $M$ defined by $\mathbb{R}^3$.

Answer 1

Iterated tensor products and wedge products are linear in each of their arguments (actually even linear over smooth functions). Thus the give zero, whenever one of their entries is zero.

Answer 2

Given $V,W$ vector spaces, we want to glue them in a manner that each pair of linear functionals $f\in V^*,g\in W^*$ raises a bilinear functional on $f\otimes g$ on $V\times W$. Addition doesn't work

$$ (f+g)(a,b+c) = f(a) + g(b) + g(c) $$

But you can check that the tensor product does, canonically,

$$ (f\otimes g)(a,b) = f(a)\cdot g(b) $$

It's pretty straigthforward to check that, given a basis $\{e^1,...,e^n\}$ for $V^*$, the set $ \{ e^i\otimes e^j\}$ is a basis for the space of bilinear functions on V. Also, can you see the answer to your second question?

Let $g:\Gamma(TM)\times\Gamma(TM)\to C^{\infty}(M)$ be a differentiable map that, pointwisely, is a bilinear map $g_p:TM_p\times TM_p\to \mathbb{R}$. Any coordinate system for $TM_p$ induces a basis $\{dx^1, ..., dx^n\}$for $T^*M_p$, by choosing

$$ dx^i\left(\dfrac{\partial}{\partial x^j}\Bigg|_p\right) = \delta_j^i $$

With this, we can work with $g$ in coordinates, at least locally, because of the remark made above, and do computations. This result is really neat, as many tools used on manifolds are multilinear maps, like $g$, and we have to do computations on them.

The second operation you mentioned is the exterior product, a way of creating an antisymmetric bilinear operation, defined as

$$ f \wedge g = \frac{1}{2}\left(f\otimes g - g\otimes f\right) $$

Finally, can you see that the steps taken above can be extended to multilinear functionals on $V$, or even in $V^*$ (because $V^{**} \simeq V$, and hit the same results? You would just replace "bilinear" with "multilinear" everywhere, and work with many more indices.