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

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?

• Note that, for any complex number $z$ and for any rank-$(m,n)$ tensor $T$, the tensor $z\otimes T=zT$ is of rank $(m,n)$ as well. Thus, if we assign a complex number to be of rank $(a,b)$, then we must have $(a+m,b+n)=(m,n)$, noting that the tensor product of tensors of rank $(p,q)$ and $(r,s)$ produces a tensor of rank $(p+r,q+s)$. Hence, $a=0$ and $b=0$. This is hopefully a good way to explain. Commented Aug 19, 2018 at 0:25
• Isn't a number just something that takes nothing and produces a number? Commented Aug 19, 2018 at 0:46
• Thank you for the reply but I don't understand your explanation! Let me rephrase my question: how/why are rank (0,0) tensors considered scalars? By definition, a tensor is formed through the combination of vectors and covectors by the tensor product and the tensor acts on vector/covector inputs to produce real numbers. What then does a rank (0,0) tensor even look like and how can it produce anything? Commented Aug 19, 2018 at 0:47
• Look at it like this. You 'feed' a rank $(n,m)$ tensor a vector and you get a rank $(n,m-1)$ tensor. You feed it a covector and you get a rank $(n-1,m)$ tensor. You can keep going and feed it $m$ vectors and $n$ covectors, and what do you get? A rank $(0,0)$ tensor, right? Commented Aug 19, 2018 at 1:01
• Huh? I guess I'm having trouble stating my question. Thanks for the answers anyways! Commented Aug 19, 2018 at 2:22

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.

• Thanks makes more sense now. So if "a" is arbitrary, then what stops us from calling a=1 all the time? Is there any practical reason why it would be anything else? What if a=0, is that a unique tensor? Commented Aug 20, 2018 at 1:43
• @RobbieFresh What $a$ is doesn't matter. Like, it really doesn't matter. The empty cartesian product is an arbitrary one-element set, so $a$ doesn't even need to be a number... it can be anything that can be an element of a set. All of these one element sets are 'the same thing' in how the relate to other sets through mappings. For instance the key property we looked at here was that any mapping $\{a\} \to X$ where $X$ is a set just corresponds to an element of $X$ (whatever element $a$ maps to). What $a$ is never even comes up. Commented Aug 20, 2018 at 2:03
• @RobbieFresh I think maybe all the whatsits here obscured what I think the message should be. Whatever framework we use to define what a $(0,0)$ tensor is, it should be a scalar. A $(n,m)$ tensor is a multilinear map that takes $m$ vectors and $n$ covectors and produces a scalar. A $(0,0)$ tensor is a map that takes nothing and produces a scalar. Something that takes nothing and produces a scalar is a scalar. Just like a map $f:\{a\}\to \mathbb R$ from a one element set to a set of scalars is a scalar. Commented Aug 20, 2018 at 2:13
• Ok, so based on what you and @Michael Albanese say, a (0,0) tensor is a map of the form T(x)=ax where a is an arbitrary scalar? Commented Aug 20, 2018 at 3:27
• @RobbieFresh Not the same $a$ as in what I was talking about (now I think I see the origin of your confusion there), but sure, that's certainly one way of looking at it. Tensors are multifacted things. Is a $(1,1)$ tensor a bilinear map that takes a vector and a covector and produces a scalar, or is it a linear map that takes a vector and produces another vector? Is a covector a linear map that takes a vector and produces a scalar, or is it a vector and actually the vectors are the covectors? (Ok that example was a little too cheech and chong). Commented Aug 20, 2018 at 3:36

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: $$a.$$

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.

• Is your statement "recall that the empty tensor product of vector spaces is the base field" an axiom that we've decided on? This is what I'm stuck on. Why is this statement true? Commented Aug 19, 2018 at 2:20
• spaceisdarkgreen's answer addresses this point. Commented Aug 19, 2018 at 11:16

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.