Functions, Functionals and Tensors

Functions.

A function is a very basic concept in mathematics, but being basic doesn’t mean it is not a source of confusion. The best place to learn what a function is, at least the best that I know of is [Ten85]. That book is not about functions per se, but a one on ordinary differential equations that starts from the very beginning and develops into a huge, complete book on the matter. If you have not access to that book (although it is cheap and I fully recommend it, because you will always need a good book on differential equations anyway), take a look for the moment on the link by Wikipedia above to clear up a bit your memory or knowledge about functions.

We will need the notion of functions because we will need the notion of a functional. And we will need the notion of a functional because (if you have been following our previous posts) we have been investing our time and energy to first clear up the notion of a tensor as much as possible. (And why is that? Well, because cleaning up concepts is one of the philosophies that drives this blog!)

Functionals.

You may start thinking of a functional as a function of a function. That is, a function that takes as its argument a function and results in a number (a scalar function). A good physics example can be found in the variational principle formulation of classical mechanics of point particles: the action functional, $S[\mathbf{q}(t)],$ that is, it takes as input a function of time — the generalized coordinate of a particle of the system at a given time, $\mathbf{q}(t).$ This function, chosen amongst all possible ones, is that which minimizes the action, and gives therefore the actual trajectory of the particle between two instants of time.

One interesting point here is that the argument of a functional, that is, a given function, can also be considered itself a vector, if that function is taken from a function space that happens to comply with the properties of a vector space (recall the axioms that define vector spaces). So a functional can be a function of a vector. Or of a set of vectors.

Let us further talk about a particular kind of functional, a linear functional, that is, one which satisfies the usual linearity requirement:

$f(\alpha \vec{v}_1 + \beta \vec{v}_2) = \alpha f(\vec{v}_1) + \beta f(\vec{v}_2),$

where $\alpha$ and $\beta$ are arbitrary real numbers, and $\vec{v}_1$ and $\vec{v}_2$ are arbitrary vectors. That is, it doesn’t matter whether you first multiply vectors by numbers and sum them, and only then evaluate the functional of that result, or evaluate first the functional of each vector and then multiply each functional evaluation by the numbers. The result is the same, a real number or scalar.

As an example, take the following linear functional:

$f(\vec{v}) = \vec{A} \cdot \vec{v},$

where $\vec{A}$ is a fixed vector and the operation above is the scalar product of vectors. You can immediately see that $f$ is a functional and that it is linear.

Now a very important point that we want to make is summarized in the representation theorem for linear functionals (see our Theorem page at the side bar), which states that if $f$ is a linear functional, then there is a unique vector $\vec{A}$ such that $f(\vec{v}) = \vec{A} \cdot \vec{v}$ for all $\vec{v}$ . All linear functionals can be precisely put into this form!

What does it mean? It states that any linear functional of a certain vector can be written as the dot product of that vector with another fixed, unique vector, which can be found appropriately. But it means much more! You can of course generalize that for as many vector arguments you want, and you will get a multilinear functional. That gives some spin-offs, and one of them certainly regards tensors.

Tensors.

So now, let us bring into the discussion the notion of a tensor, from the point of view of “simple tensors” (we will generalize that further in future posts). They are simply defined as linear vector-valued functions of one of more vectors (that is, being linear in their arguments). That is, functions that maps vectors to a vector in a linear way.

Functionals and Tensors.

So we have at this point a pretty abstract definition of a tensor (a linear vector-valued functions of one of more vectors) at one side and a multilinear functional (a scalar function of functions, which in turn can be vectors) at the other side, but then you may be asking what is the connection between them?

Well, it is in fact simple to see the connection if you have really understood the implications of the representation theorem for linear functionals and have a good idea on how to apply them to get pretty interesting results. The clue is: through that theorem you can construct linear vector-valued functions of vectors (i.e., tensors) directly from multilinear functionals and vice-versa, and in fact generalize the notion of tensors, making a unified picture in which a tensor is a multilinear functional itself.

To see that, start with a bilinear vector-valued function of two vectors, $f(\vec{v}_1,\vec{v}_2)$ . Hold, say, $\vec{v}_2$ fixed for a moment. Then under that condition it is clear that $f(\vec{v}_1,\vec{v}_2)$ defines a linear functional on the variable $\vec{v}_1$ , right? So if you apply the representation theorem for linear functionals to $f(\vec{v}_1,\vec{v}_2)$ , when $\vec{v}_2$ is held fixed, you may write that:

$f(\vec{v}_1,\vec{v}_2) = \vec{A} \cdot \vec{v}_1.$

Notice, however, that since we have temporarily fixed $\vec{v}_2$ , $\vec{A}$ depends linearly on $\vec{v}_2$ . Then what we are saying is that $\vec{A}$ defines a new linear vector-valued function of a vector, which we will call $T$ :

$\vec{A}= T(\vec{v}_2).$ .

But if that is linear vector-valued function of a vector, then it defines a tensor.

So, you see, that game is also true if you had started from the tensor $\vec{A}$ and wanted to get the bilinear functional $f(\vec{v}_1,\vec{v}_2)$ . You can also generalize that to multilinear functionals as well as vector-valued functions of several variables in one direction or another.

There is then a close association between tensors (as strictly defined as linear vector-valued function of vectors) and linear functional — back and forth — through the representation theorem for linear functionals. It is very natural that we want to extend the notion of a tensor to include multilinear functionals.

The rank of a tensor is the sum of the number of arguments and the nature of the output. So if you have a tensor that maps three vectors and as gives as output one vector, then it is a tensor of rank 4.

Then, in that previous example, the tensor $\vec{A}$ is a linear functional that takes as its arguments 1 vector and delivers 1 vector, the so-called tensor of rank 2 — it maps a vector to a vector (can you imagine a map that rotates a vector through an angle?). A tensor $f(\vec{v}_1,\vec{v}_2)$ is of rank 2, that is, it maps 2 vectors to a scalar (that scalar “counts” as a zero); recall the metric function $d(P_1,P_2)$ ? Notice that a linear scalar function (one that takes as argument a scalar and delivers a scalar) is a tensor of rank 0.

You can play around with tensors so redefined, like, e.g., a tensor that gets as argument, say, X vectors and delivers, say, Y vectors.

Generalizations.

As a final word, recall that we are talking about Euclidean tensors. We will generalize that notion to tensors living on more general spaces later. Then it will happen that a tensor will take as arguments not only vectors, but also other geometrical objects (one-forms), so that the inputs and outputs result in completely “mixed” tensors, with input geometrical objects living in different spaces. More tensor analysis on manifolds later.

So for the moment, a scalar is a tensor of rank 0. A vector is a tensor of rank 1. A tensor can be defined through a multilinear functional.

Ref.: [Nea10]

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Written by Christine

January 3, 2011 at 10:47 AM

Posted in Mathematics, Tensors

Tagged with Basic

18 Responses

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Quickly edited to include subsections and a few clarifications.

Christine

ccdantas

January 3, 2011 at 11:16 AM

Reply
Hi Christine,

So I see you have dedicated at least part of your vacation time to continuing the development of this exploratory blog. I assume once you’ve mapped out the essentials you will move on to examine some of the particulars they relate to. I realize what you are reviewing in this post must present as so basic its almost intuitive, yet as my mathematical background being more piece meal than anything else, I need to pay particular attention here as to attempt having the concepts straight, even if my fluidity with them remains awkward. However so far I have not found any physical concepts with which the math(s) relate I’m not familiar with. So thanks and as I said in the outset I’m curious as to where all this will lead as time progresses.

“I admire the elegance of your method of computation; it must be nice to ride through these fields upon the horse of true mathematics while the like of us have to make our way laboriously on foot.”

-Albert Einstein, taken from correspondence with Tullio Levi-Civita (Mathematican), during the final stages of developing General Relativity.

Best,

Phil

Phil Warnell

January 3, 2011 at 2:54 PM

Reply
Hi Phil,

Well, a good part of that post was already written back in the end of last year… So it was a question of just wrapping things up a bit.

Yet… although I am in vacation, my travel has been cancelled, as my husband is at home due to a bike injury that he suffered in the Dec 30th (you may check recent twits). My mother is also here (she came for the end of the year festivities), but she could not return to her home town yet due to my husband’s accident (she is elderly and cannot travel alone). So I had a lot of things to solve these last few days, like going back and forth the hospital, arranging documents and all, cooking, taking care of home and everything.

I am exhausted! Yet, if I find energy and time, I’ll be posting in January as well, since I’ll be home anyway. But I cannot make guarantees.

I hope that my development in this blog leads to something useful that can be added to your knowledge. For the moment, it’s all a work in progress, and in small bits. (Maybe that will be just too boring for those who enjoy a fast pace). And yes, I am currently focusing on basic material.

Best,

Christine

ccdantas

January 3, 2011 at 4:30 PM

Reply
Hi Christine,

Why does it seem that this time of year comes so often with mixed blessings? That’s to have you know I’m sorry to hear about your husband’s accident and the need to cancel your travel plans. I hope things are not too serious with him and somehow you both, along with your son, will manage to get some pleasure from what holidays remain.

As for the blog I hope that I haven’t given you the impression I have any level of prior expectation, for no matter at what rate produced or how far you might take I find it having value. For that matter I find anytime one is able pull thoughts out of their head to write them down it has its benefits, if only as it refers to self examination respective of their soundness if nothing else. So first things first, is to get some needed rest and have your husband to recover, with what comes after to be what it will be.

Best,

Phil

Phil Warnell

January 3, 2011 at 5:09 PM

Reply
Hi Phil,

Thanks for the kind words! Yes, although the bike accident came unexpectedly (as by definition any accident), and resulted into several spin-offs, everything seems to be progressing relatively fine now, since he is now under a treatment at home that might avoid surgery.

And it’s nice to know that you have grasped the overall idea of this blog, thanks!

Christine

ccdantas

January 3, 2011 at 5:28 PM

Reply
So for the moment, a scalar is a tensor of rank 0. A vector is a tensor of rank 1. A tensor can be defined through a multilinear functional.

And Mechanical Engineers such as myself are taught in our undergraduate days when we study 3×3 matrices in Statics (Stress and Strain) (still Euclidean), such a Matrix is a tensor of rank 2.

I imagine we will get to the 4 x 4 matrices so vitally important in Gen Rev. This is correct, yes? Or have I missed the point?

Steven Colyer

January 5, 2011 at 2:38 PM

Reply
Also, my “get well” wishes to your husband. I once slipped on a puddle of juice from my 4th child’s sippy cup, fell down the stairs and broke my ankle in 3 places. My bad for not cleaning his mess up. Six months later I could walk again normally. Not quite as macho as your husband’s accident. But broken bones do heal in time, thank goodness.

Very nicely done re this article, Christine. You have a natural aptitude for good expository teaching. We’re learning much, grazi.

Steven Colyer

January 5, 2011 at 2:50 PM

Reply
- Thanks for your best wishes; it was not a broken bone but the clavicle went out of place…
  
  And also thanks for the nice word on my article. As we progress, I hope things get clearer and clearer.
  
  Re: matrices, see my comment here, just posted.
  
  Best,
  Christine
  
  ccdantas
  
  January 5, 2011 at 3:03 PM
  
  Reply
Dear Steven,

Yes, you missed the point.

A matrix is just a *representation* of a tensor. It is not a definiton of a tensor. As an engineer, you have used matrices to represent tensors in order to make calculations with them. That means you choose a basis to construct these matrices. We will use matrices eventually, yes, but also, and more importantly, abstract concepts of tensors in which we will not fix any representation at all, and see them more as geometrical objects that live independently of any representation or coordinate system.

I am aware that this blog may possibly bore some readers that just want to grasp tools to be promptly used for pragmatic calculations and run fast from point A to point B. The idea of this blog is not to get to point B as fast as possible, but to develop concepts and equally question the whole process between A and B. That will take as long as necessary, and I might even not be able to reach point B at all, since that is the end and the end does not necessarily hold all the answers to the questions we initially made. The journey is what counts.

So if you are looking for a quick and fast review of GR, you will certainly find a lot of good material elsewhere available freely in the internet. Here, the philosophy is another one.

Hope that clarification helps.

Best,
Christine

ccdantas

January 5, 2011 at 3:00 PM

Reply
[...] that reason I have decided to write this short note, motivated by a reader with an engineer background (see a brief clarification in a subsequent comment). To relief that [...]

A note on matrix representation of tensors « Toy Universes

January 5, 2011 at 5:31 PM

Reply
Ah, thank you, Christine, that cleared things up on this important subject nicely.

I see where my confusion came from it was from the following line ….

Objects that transform like zeroth-rank tensors are called scalars, those that transform like first-rank tensors are called vectors, and those that transform like second-rank tensors are called matrices.

.. in Eric W. Weisstein’s page at Mathematica re Tensors: here.

Steven Colyer

January 5, 2011 at 8:13 PM

Reply
I meant: here.

Steven Colyer

January 5, 2011 at 8:15 PM

Reply
Hi Steven,

That’s odd and confusing; I’d would not put it that way! A scalar could perfectly be a one by one matrix, and a (column) vector, a n by one matrix, where n is the dimension of the vector space. I mean, these could all be regarded as matrices.

Maybe they were raising a terminology issue specific to computer science? I’m not sure.

Best,
Christine

Edit: Better saying, a scalar can be represented by a 1 x 1 matrix; a vector, by a n x 1 matrix, etc.

ccdantas

January 5, 2011 at 8:31 PM

Reply
And BTW, see the newest post of this blog.

ccdantas

January 5, 2011 at 8:43 PM

Reply
[...] tensors. Recall from previous posts that we have extended the definition of tensors to include multilinear functionals. So the picture we have up to now is that a tensor is like a machine with a certain number of slots [...]

Tensors in terms of components « Toy Universes

January 6, 2011 at 10:00 PM

Reply
[...] that aim, if you recall the representation theorem for linear functionals (introduced in this post; see also the Theorems page in the side bar), you know that the bilinear functional above is [...]

Tensors in terms of components II: changing basis « Toy Universes

January 7, 2011 at 4:57 PM

Reply
[...] 1) a multilinear functional: [...]

More details on two views of tensors « Toy Universes

January 12, 2011 at 6:41 PM

Reply
[...] a comment » When discussing functionals and their relation to tensors for the first time, in this post, we mentioned the representation theorem for linear functionals (see also our Theorems page at [...]

Representation theorem for linear functionals in component form « Toy Universes

February 9, 2011 at 10:22 AM

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