How is a (0,0) rank tensor a number?

Question

I've always been confused by this. If tensors are basically functions whose inputs are vectors/convectors (tensors) and outputs are numbers (or other nth-ranked tensors), then how does a tensor that cannot accept input arguments produce a number? Can someone give me a concrete example of how this works please?

Answer 1

Let $V$ be a real vector space. A $(p, q)$ tensor is a linear map $T : (V^*)^{\otimes p}\otimes V^{\otimes q} \to \mathbb{R}$. Now recall that the empty tensor product of vector spaces is the base field, so if $(p, q) = (0, 0)$, we have $(V^*)^p\otimes V^q \cong \mathbb{R}$. Therefore a $(0, 0)$ tensor is a linear map $T : \mathbb{R} \to \mathbb{R}$. Such a map is necessarily of the form $T(x) = ax$ for some $a \in \mathbb{R}$, namely $a = T(1)$. This provides us with a one-to-one correspondence between $(0, 0)$ tensors and real numbers.

Answer 2

Does it help if you think of it in terms of coordinates?

A type $(m,n)$ tensor has $m$ indices upstairs and $n$ downstairs: $$ a^{i_1 \dots i_m}_{j_1 \dots j_n} . $$ So a type $(0,0)$ tensor has no indices: $$ a. $$

Answer 3

You mentioned in a comment that you thought understanding why the nullary tensor product is the one-dimensional space was the crux of the question. Of course this can simply be an ad-hoc definition for a degenerate case, but it's important to see why it's natural.

If we're avoiding defining tensor product spaces and just defining an $(n,m)$ tensor as a multilinear map $V^n\times (V^*)^m\to \mathbb R$ then the $(0,0)$ case is a multilinear map from the nullary cartesian product to $\mathbb R.$ The nullary cartesian product is an (arbitrary) one-element set $\{a\}.$ And maps $\{a\}\to \mathbb R$ are in obvious one-to-one correspondence with $\mathbb R.$

This is just a roundabout way getting at the intuitive argument I gave in the comments earlier (and I'll put it more eloquently here): "a function that takes no arguments and spits out a real number is a real number."

You could just as well ask why the nullary cartesian product is a one-element set, but my answer (without resorting to category theory) would just be that is the definition that makes the above intuition about functions with no arguments work out. I could give the answer that resorts to category theory, but I think I'll save the category theory for fleshing out the nullary tensor product.

For this we have to think about what a tensor product is. It takes vector spaces and makes a new vector space. For the binary tensor product, we can write it like $$f:V_1\times V_2\to V_1\otimes V_2.$$ The key property is that $f$ is in some sense the most general bilinear map. More formally, we have the universal property that for any vector space $W$ and bilinear map $g: V_1\times V_2 \to W,$ there is a unique linear map $\tilde g:V_1\otimes V_2\to W$ such that $g = \tilde g \circ f.$ (See the commuting diagram at the link.) The fact that every bilinear map out of $V_1\times V_2$ can be instead thought of as a linear map out of the tensor product space is what characterizes the tensor product space.

Now if we take the nullary case of this, we need a vector space $\otimes^0V$ and a map $$f:\times^0 V\to \otimes^0V$$ where recall $\times^0 V$ is some arbitrary one-element set $\{a\}.$ Now consider an arbitrary vector space $W$ and a map $g:\{a\} \to W,$ which is, again, really just an element of $W.$ Let's call it $w$ We need to find a vector space $\otimes^0V$ and map $f:\{a\} \to \otimes^0V$ such that there is a unique linear map $\tilde g: \otimes^0V \to W$ with $g=\tilde g\circ f.$ Of course we already know what the vector space should be: $\otimes^0V = \mathbb R.$ For $f,$ it turns out we can choose any nonzero element $r\in\mathbb R$: let's take $r=1$ for simplicity.

Now we can see that there is indeed a unique $\tilde g$ with the required properties: the linear map $\mathbb R\to W$ that takes $1\mapsto w.$ Thus $\mathbb R$ has the universal property for the nullary tensor product.

Whether this generalization from binary to nullary via the universal property really justifies the definition of the nullary product as $\mathbb R$ is debatable. I tend to think of it more of an endorsement of the universal property approach that it gets the corner case 'correct', i.e. it agrees with the intuitive arguments that lead one to conclude that the $(0,0)$ tensors are scalars and thus the nullary product tensor product space is one-dimensional.

Answer 4

Concrete Examples:

Take the $(0,0)$-rank tensor $T=1$. Then taking zero vectors and zero convectors, it produces the number $1\in\mathbb R$.

Also, recall that the exterior differential operator $\mathrm d$ takes in an anti-symmetric $(0,p)$-tensor and produces a $(0,p+1)$ tensor. It is defined as the antisymmetrized partial derivative:

$$ \mathrm (d\omega)_{\mu_1\cdots\mu_2\nu}=\partial_{[\nu}\omega_{\mu_1\cdots\mu_2]} $$

So, a scalar can be turned into a $(0,1)$ tensor, namely the gradient. Therefore a scalar must be a rank $(0,1-1)=(0,0)$ tensor.