I've been studying multilinear algebra on Kostrikin's "Linear Algebra and Geometry" and he says the following. If $V$ is a linear space, $T^q_0(V)=V^{\otimes q}$ and if $f_\sigma :T^{q}_0(V)\to T^q_0(V)$ is given by:

$$f_\sigma(v_1\otimes\cdots\otimes v_q)=v_{\sigma(1)}\otimes\cdots \otimes v_{\sigma(q)}$$

Then the subspace of symmetric tensors is generated by the map

$$\frac{1}{q!}\sum_{\sigma\in S_q}f_\sigma:T^q_0(V) \to T^q_0(V)$$

I understood his proof of this. However I can't see the intuition behind it. What's the motivation of summing all those functions and why we must include that factor $1/q!$ it's what I don't understand. The same happens with the alternation operator, which he defines as:

$$\frac{1}{q!}\sum_{\sigma \in S_q}\operatorname{sgn}(\sigma)f_\sigma : T^q_0(V) \to T^q_0(V)$$

Again he proves that the image of this things is really the space of antysimmetric tensors. However again, what's the intuition behind this? How can we see intuitively that any symmetric tensor can be written as an arbitrary tensor under the action of this function. And again, I can't understand why this factor with $q!$ comes again.

I believe there is some intuition, although the author doesn't tell. It's like the tensor product, the intuition is that we want to construct a pair $(S,T)$ of a vector space $S$ and a multilinear function $T$ such that the pair possess the universal property. So we do all that job with the free vector space and there are good motivations, good reasons to construct things like they are. It's something that after we read about the motivations and so on we can think: "those are good reason, I can see why someone ever thought of them!". I believe there must be good reasons too for the introduction of these operators as they are.

How can we see intuitively that it must be like that? Can someone help with this?

Thanks very much in advance!


For the start let's have a look at matrices.

Let $A \in \mathbb{R^{n\times n }}$ be any matrix.

Than $\text{sym}(A)=\frac{1}{2}(A+A^T)$ is it's symetric part and $\text{skew}(A)=\frac{1}{2}(A-A^T)$ its antisymetric part. Note that any matrix can be written as sum of its symetric and antisymetric part $A=\text{sym}(A) + \text{skew}(A)$

Now I explain why there is the factor $\frac{1}{2}$. If matrix $A$ is already symetric you want to have $\text{sym}(A) = A$. Without the factor you would get $\text{sym}(A) = 2A$. Similar for antisymetric part.

You can think about matrices as bilinear forms, $x^T A y = A(x,y)$. Matrix $A$ is symetric iff $A(x,y)=A(y,x)$ for all $x,y$. This brings as to idea to what symetric tensor might be. That $T$ multilinear form(or tensor) is symetric iff $T(...,x,...,y,...) = T(...,y,...,x,...)$ for all $x,y$ and for all positions where you preform the 'swap'. As with matrices you can take symetric part of tensor $T$:

$$\text{sym}(T)(x_1,...,x_n) = \frac{1}{n!} \sum_{\sigma\in S_n} T(x_{\sigma(1)},..,x_{\sigma(n)})$$

Again natural requirement is if you have symetric tensor $T$ than $\text{sym}( T)=T$, this explains the factor $\frac{1}{n!}$

And why bother with symetric and antisymetric tensors? Well I'll give few remarks from my studium.

In physics you encounter tensors quite a lot and they are often symetric or anytisymetric, like electromagnetic tensor is antisymmetric. Or in general relativity you often caclulate something like this $\sum_{ij} A_{ij}B_{ij}$. When $A$ is symmetric than $\sum_{ij} A_{ij}B_{ij}=\sum_{ij} A_{ij}\text{sym}(B)_{ij}$ which is useful.

Or in differential geometry you define differential forms which are antisymetric and thay are of great importance.

I hope I shed a little bit of light on this topic.


I've been trying to wrap my head around "differential forms" and tensors for some time now, longer than I'd like to admit. I've been reading from three different texts (primarily), those being: "Differential Geometry of Curves and Surfaces" by Docarmo, "Differential Forms with Applications to the Physical Sciences" by Flanders, and "Calculus on Manifolds" by Spivak. It hasn't been until I've recently decided to be disciplined and really eat Spivaks manifolds that I've really been able to pin anything useful down.

I think I'm like you, when I learn something, I don't just want to know it, I want to know how it was discovered, how to discover it on my own. According to LaToza and Myers in "Hard-to-Answer Questions about Code," "rationale" is the most difficult thing to understand when trying to comprehend a computer program. Indeed, what things do and why they work are fairly "straightforward," but "why was this program written this way" is almost impossible to know without some understanding of the thought process of who wrote it (that's almost a tautology). Indeed, some contextual information or insight into the problem that was being solved is pretty critical for that piece of the pie.

I don't know what the originators of the anti-symmetric tensor were thinking when they invented this beast, but I'll give a shot at "deducing" it from a reasonable attempt to solve a problem here.

So, let's start with trying to determine the volume of a parallelipiped (please don't ever ask me to pronounce or spell that). You know, the volume determined by some vectors... Anyways, let's suppose that there is a function that can determine this volume (seems reasonable). What should the dependent variables be? Well, it should certainly depend on every vector. It is "reasonable" to suppose it does not depend on anything else... For example, if we calculate its volume today, it should not be different tomorrow - this is something special about $R^n$, but it's not too surprising (I mean, why should, in our universe, volume remain constant?).

So, we have a function of our vectors, and nothing else. Let's swallow the next pill, multi-linearity. This is essentially Cavalieri's principle, but I'll elaborate a little bit. What happens if we scale just one of our vectors? Well, think of all the other vectors as forming a "face," and this vector we're scaling as determining "thickness." By scaling a vector we're scaling the thickness, and should sort of scale the volume. This isn't such a hard pill to swallow, but the "additive" part of linearity is a bit funnier. I recommend that you draw a parallelogram on some x-y axis, and "add" a vector to the "upper" vector, and see what the area will come to. Approximate by (arbitrarily) thin rectangles if it helps, and refer to some nicer illustration of Cavalieri's principle.

(Or meditate on the following ascii art if it's enough)

.... .... .... -> .... .... ....

So, at this point we "agree" that determining our volume should be a multi-linear, real valued function of our vectors? That's precisely what is called a "tensor," in our case we're dealing with an "n-tensor" in $R^n$. The hard, if not just plain weird part, is coming up with this anti-symmetric stuff (sorry for breaking character: we are reverse-engineering this thing). I guess the first, ominous, thing to note is that by accepting multi-linearity we've also accepted "negative volumes." What might be more useful is to consider first what should have 0 volume. I say that it is absolutely necessary, in our fake world of $R^n$, that if two vectors of our parallepiped are the same, then it should have volume 0. I mean, in this case it is "flat" in (at least) one dimension, and if you don't call that 0 volume then I quit. We shall see that this concession, and that of multi-linearity, has essentially sealed our fate.

Let's call our volume tensor $T$, and consider $R^2$. We'll take two respectable (i.e. linearly independent, non-zero) vectors $u$ and $v$. Clearly, $T(u+v,u+v)$ should be 0, as this is the area of one single vector (as we've previously discussed). Given that we've accepted multi-linearity, we end up with: $$ 0 = T(u+v,u+v) = T(u,u) + T(v,v) + T(u,v) + T(v,u)$$ The two terms on the left are again 0, and so the two on the right must be each other's negative. So, this hints at "anti-symmetry," but is it sufficient to get our desired property of "flat has volume 0"? Well, yes, because:

$$T(..., u_i, ..., u_i, ...) = -T(..., u_i, ..., u_i, ...) \Rightarrow$$ $$2T(..., u_i, ..., u_i, ...) = 0$$

Note that in the first expression I exchanged the position the first $u_i$ and the second $u_i$... (that's a joke...)

As a side note, using multi-linearity we see that indeed the arguments must be linearly independent to achieve non-0 volume, which is the appropriate formalization of (i.e. scholarly way of saying) "not-flat."

Now, as is natural in math, once sufficiency has been determined, one asks of necessity. Does our "volume tensor" have to be anti-symmetric? There are A LOT (I mean, a plethora, buttload, $n^n$, etc.) of tensors that one could come up with, so it's not obvious that ONLY the anti-symmetric ones make sense. Well...

Let's suppose that we have a $u$ and a $v$ such that: $$ T(..., u, ..., v, ...) + T(..., v, ..., u, ...) \neq 0 $$ Let's then calculate: $$ 0 = T(..., u + v, ..., u + v, ...) = \\ T(..., u, ..., u, ...) + T(..., v, ..., v, ...) + T(..., u, ..., v, ...) + T(..., v, ..., u, ...)$$

Now, the only way this makes sense, since the two terms on the right don't sum to 0, is if either (or both) of $T(...,u,...,u...)$ and $T(...,v,...,v,...)$ aren't 0, which violates one of our presumptions about our tensor. This means that our tensor must be anti-symmetric in order to be our "volume-tensor."

There is some decent exercise in finding out that the anti-symmetric tensors are intrinsically related to the classical determinant (the only degree of freedom is the tensor's value at the basis), and this answer is already long, so I won't do it here (hint: the vector space of anti-symmetric n-tensors in $R^n$ is one-dimensional). But, I would like to make a comment that this final line of thought was rather "algebraic" and less "geometric" than I'd have liked it to be. I think it is sort of the clearest and most direct path from a to b, but going backwards (even more backwards) we can come up with a "geometric" thought process to get us to anti-symmetry.

Remember back when I was convincing you to accept multi-linearity, then I mentioned that we must also accept negative volumes? Well, let's suppose another geometrically plausible position, that volume is invariant under rotation. Let's go back to $R^2$, specifically our $u,v$. We've already agreed that $T(-u,v) = -T(u,v)$, now lets rotate those bad-boys -90 degrees... $$R_{-90}(-u,v) = [[0,1],[-1,0]][-u,v] = (v,u)$$ And there we have it, $T(v,u) = T(-u,v) = -T(u,v)$, which strongly suggests investigating anti-symmetry in more generality...



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