25
$\begingroup$

So, I've been reading up on multilinear algebra a bit. In particular, I've been reading up on the construction of of the exterior algebra of a finite dimensional vector space $X$, say over $\mathbb{R}$. $$ \Lambda(X) = \bigoplus_{n \geq 0} \Lambda^k(X) $$

I'm still at that frustrating early stage where the definitions seem very unmotivated. I'm hoping for some suggestions for improving my grip on them.

Let me describe one particular thing which is bothering me in hopes that my concerns are easily dispelled. I don't understand the point of having a product $\Lambda(X) \times \Lambda(X) \to \Lambda(X)$ instead of just paying attention to the products $\Lambda^k \times \Lambda^\ell(X) \to \Lambda^{k+\ell}(X)$ which seem to be all that is important.

My problem may be that the only case I have any experience with is the case $X=X^*$ (a dual space) in which case the elements of $\Lambda^k(X^*)$ can be identified with alternating $k$-linear functionals $X^k \to \mathbb{R}$. It seems strange to me to want to put the forms of different ranks together in the same algebra. What is the use of an expression of mixed rank like $$\omega = dx + dy \wedge dz$$ which is, I suppose, an element of $\Lambda((\mathbb{R}^3)^*)$?

I think the thing which irritates me the most is that these mixed expressions do not even necessarily alternate! I mean, for Pete's sake, look! \begin{align*} \omega \wedge \omega & = (dx + dy \wedge dz) \wedge (dx + dy \wedge dz) \\ &= dx \wedge dy \wedge dz + dy \wedge dz \wedge dx \\ &= dx \wedge dy \wedge dz + (-1)^2 dx\wedge dy \wedge dz \\ &= 2 dx \wedge dy \wedge dz \\ &\neq 0 \end{align*} What's the point of considering all these extraneous elements whose wedge square isn't even zero?

Now one answer to my question might be "well, isn't it useful to consider polynomials which aren't of homogenuous degree?". I don't think this is good enough for me though. Until I see why it is really useful to put the "$\Lambda^k(X)$"s together into an algebra, I'm going to be wary of the object $\Lambda(X)$.

Added: I noticed there is some relevant information at this thread.

$\endgroup$
  • 3
    $\begingroup$ Determinants are antisymmetric multilinear forms and exterior algebra is a way to formalize the properties of determinats and their minors. $\endgroup$ – Emilio Novati Mar 10 '15 at 20:14
  • $\begingroup$ @EmilioNovati: Your remark about determinant minors sounds interesting. I'll look into this. $\endgroup$ – Mike F Mar 10 '15 at 20:21
  • 5
    $\begingroup$ Important: The equality $\omega\wedge\omega=0$ does not hold in general, even when considering only homogeneous forms. For example, consider the standard symplectic form on $\mathbb{R}^{2n}$. $\endgroup$ – Amitai Yuval Mar 10 '15 at 20:26
  • 1
    $\begingroup$ @MikeF: And note that determinants are volumes and minors are areas of the faces. So forms are oriented areas or volumes. $\endgroup$ – Emilio Novati Mar 10 '15 at 20:39
  • 1
    $\begingroup$ I am tempted to give the following application: exterior algebras are important because they are particular cases of Clifford algebras which are important in full, not just their homogeneous coordinates, for example in constructing the Atiyah-Bott-Shapiro isomorphism (aka algebraic Bott periodicity). But I'm not comfortable enough with these concepts to post this as an answer, and it is stretching it a bit (ABS theorem does not consider trivial Clifford algebras, i.e. exterior algebras). $\endgroup$ – Bruno Stonek Mar 11 '15 at 16:49
9
$\begingroup$

You might prefer to think of the exterior algebra as a graded algebra. A graded algebra is a monoid in the monoidal category of graded vector spaces, just as an algebra is a monoid in the monoidal category of vector spaces.

Wikipedia defines these objects as algebras (respectively, vector spaces) with extra structure, but you don't have to. You get equivalent categories with the following definitions. A graded vector space is a sequence $(V_n)_{n \in \mathbb{N}}$ of vector spaces -- the sum of elements in different grades is not taken to be well-defined. A morphism of graded vector spaces is a sequence $(f_n)_{n \in \mathbb{N}}$ of linear maps, and the tensor is given by $(V_n)_{n\in \mathbb{N}} \otimes (W_m)_{m \in \mathbb{N}} = (\oplus_{n+m=p} V_n \otimes W_m)_{p \in \mathbb{N}}$. A monoid in this monoidal category turns out to be exactly what you describe: a sequence of vector spaces $(A_n)_{n \in \mathbb{N}}$ with multiplication maps $V_n \otimes V_m \to V_{n+m}$ satisfying unit and associativity laws.

(There are actually two important ways to make this into a symmetric monoidal category: one symmetry is $\sigma (v_n \otimes w_m) = w_m \otimes v_n$ while the other is $\sigma(v_n \otimes w_m) = (-1)^{nm} w_m \otimes v_n$. Under the second symmetry, the exterior algebra is actually a commutative monoid!)

Anyway, the point is that the exterior algebra is a graded algebra, and the category of graded algebras can be defined in different ways. In some of these ways, we think of the sum of elements in different grades as being well-defined, while in others we don't. It's a matter of taste.

EDIT I can't resist pointing out that "taking the sums of elements of different grades to be well-defined" amounts to defining a functor $\mathsf{GrVect} \to \mathsf{Vect}$ from graded vector spaces to vector spaces, sending $(V_n)_{n \in \mathbb{N}} \mapsto \oplus_n V_n$. But this isn't even the only useful such functor -- another one would be $(V_n)_{n \in \mathbb{N}} \mapsto \Pi_n V_n$, which differs when there are infinitely many nonzero grades. Admittedly this functor doesn't play as well with the monoidal structure.

EDIT 2 The description I gave of $\mathsf{GrVect}$ essentially regards it as the functor category $[\mathbb{N}, \mathsf{Vect}]$ (where $\mathbb{N}$ is regarded as a discrete category). The monoidal product is given by Day convolution (where $\mathbb{N}$ is regarded as being monoidal under addition), a general way of producing monoidal structures on nice functor categories.

$\endgroup$
  • 4
    $\begingroup$ I sympathize with your answer, which is the same thing I thought when I read the question, but that still doesn't say why it would be useful to consider the whole graded algebra and not just its homogeneous components separately... $\endgroup$ – Bruno Stonek Mar 11 '15 at 16:40
  • $\begingroup$ Fair enough. Some of the other answers here present concepts that have nice, clean definitions in terms of the whole exterior algebra. Probably these definitions could be reproduced in the setting I outlined (and they might be comparably clean), but it's certainly nice that they have clean formulations from the whole-exterior-algebra perspective. $\endgroup$ – tcamps Mar 11 '15 at 17:25
  • 2
    $\begingroup$ @lenticcatachresis: The exterior algebra is an "odd" analogue of the symmetric algebra (which, more or less, is a generalization of the polynomial algebra). So you could just as well ask why we study (multivariate) polynomials and not just their various homogeneous components separately. Most of the multivariate polynomials people are concerned with are homogeneous, after all. But I (and many people) find it easier to think of multiplication as a graded map $A \times A \to A$ than to think of it as a family of maps $A_n \times A_m \to A_{n+m}$, for example. And the fact that graded ... $\endgroup$ – darij grinberg Mar 25 '15 at 1:51
  • $\begingroup$ ... are algebras (whence one can apply to them everything one knows about algebras, without having to reprove it with the word "graded" inserted everywhere) comes handy. $\endgroup$ – darij grinberg Mar 25 '15 at 1:51
8
$\begingroup$

Nice question! Well, let's say you just have the $\Lambda^k(T(M))$ in isolation ($M$ is a manifold, $T(M)$ is its tangent space). What are they? Modules over the $C^n(M)$ functions ring? Nice, but how can we multiply one form with another form? We can't, we have modules not rings or algebras. Yet we would like to multiply a 1-form with another 1-form to build a 2-form, for example. OK, so we invent the exterior product $\wedge^{k,s}$. It is called exterior because it takes a k-form and a s-form and gives you back a (k+s)-form. It goes out of the original modules. We actually have a collection of these exterior products, one for each pair of $k,s$. So we have a collections of modules $\Lambda^k(T(M))$ and a collection of exterior products $\wedge^{k,s}$. What a mess. When faced with similar situations mathematicians just generalize things, to make them simpler. They did exactly the same thing with polynomials after all. Think about it: there is no internal multiplication between polynomials of order $k$.

So let's create a mathematical structure, $\Lambda(T(M))$, which contains all the $\Lambda^k(T(M))$. $\Lambda(T(M))$ is the direct sum of all the $\Lambda^k(T(M))$ There is now an internal multiplication which is called the wedge product $\wedge$, which subsumes all the $\wedge^{k,s}$. There is also an internal addition and scalar multiplication which subsume all the different additions and scalar multiplications in the $\Lambda^k(T(M))$. Actually the scalar multiplication is the same as the wedge product with $\Lambda^0(T(M))$. So what we have got is an algebra! Great, we went from a collection of structures and external operations to a single mathematical structure. Much cleaner. You can also create a single derivative internal operator which subsumes all the different exterior derivative operators acting on the individual $\Lambda^k(T(M))$.

But the most important construction using the full power of your new graded algebra $\Lambda(T(M))$ is the Ideal. This construction requires the freedom to add and multiply forms with different degrees. You cannot do it with modules, you need a ring at least.

I suggest you read "Applied exterior calculus" by Edelen to see the neat things you can do with $\Lambda(T(M))$

Now, do you still find it uncomfortable to sum a 1-form with a 2-form as in your $\omega$?

It boils down to psychology.

Think about it this way: when you were a kid they showed you an apple sitting next to another apple and they told you to describe this as 2 apples. In math you wrote $1+1=2$. You asked: can I do this with bananas too? Sure, you can do this with anything and the result is always $2$, but....if you have an apple and a banana you cannot sum them anymore. You can only sum apples with apples and bananas with bananas.

Bad,Bad,Bad teacher.

Fast forward to today and you do not know what to do with 1-form sitting next to a 2-form.

But suppose you had a better teacher. She shows you 3 and a half apples and 2 bananas next to each other on a table and ask the smart kid in your soul to describe this (the smart kid is there, just look for it).

The smart kid says: oh, as you told me yesterday, it's a vector in a 2D vector space over the rationals. I would describe it as $3.5 a + 2 b$ where $a$ and $b$ are base vectors "apple" and "banana".

So the teacher asks: what if I give you another half apple and take away a banana?

The kid replys: now I have $4a+b$, it's a vector space with basis a and b, so you still sum apples with apples and bananas with bananas, but if you have apples and bananas you just imagine you have a vector space with more dimensions instead of the usual 1D space of apples or bananas (which is equivalent to a scalar, that is a simple number).

The bad teacher thought that "apples" and "bananas" were numbers. They are in fact vectors in 1D vector spaces. When you sum them together, you are doing a "direct sum" of vector spaces.

Fast forward to today. You have a 1-form sitting next to a 2-form what do you do? Just add some milk and mix yourself a great mathematical smoothie :-)

$\endgroup$
  • 1
    $\begingroup$ As an application of ideals in the exterior algebra, differential ideals appear to be relevant to differential equations. $\endgroup$ – tcamps Mar 11 '15 at 17:23
  • 3
    $\begingroup$ Note that we are perfectly okay with having different multiplications on different-sized matrices $R^{m\times n}$ - we don't try to form linear combinations of incompatible matrices - so I don't think your "what a mess" argument holds water. $\endgroup$ – Mario Carneiro Mar 18 '15 at 3:59
  • $\begingroup$ @MarioCarneiro what matrices are you referring to? I never said anything about matrices. We are talking about (graded) vector spaces or modules and why they are useful in mathematics $\endgroup$ – magma Mar 20 '15 at 1:38
  • 4
    $\begingroup$ @magma I know, I'm just giving another example where we have a similar division of natural operations into different groups, and just live with the fact that the matrix multiplication operation has a weird closure condition. My point is, what makes an exterior algebra different so that we feel the need to force 1-forms and 2-forms to live in the same vector space by constructing the free vector space generated by the different forms and ending up with "mixed forms"? Why is this okay but $\begin{matrix}1&0\\0&1\end{matrix}+\begin{matrix}1&0&0\\0&1&0\\0&0&1\end{matrix}$ is not? $\endgroup$ – Mario Carneiro Mar 20 '15 at 2:55
6
$\begingroup$

Focusing on how such mixed-grade objects could be useful, expand slightly to clifford algebra for a moment. The clifford product of two vectors produces a mixed-grade object called a rotor. For instance, using $\mathbb R^3$ as a base for a clifford algebra, with basis vectors $e_1, e_2, e_3$, then the clifford product of two vectors $a, b$ yields some linear combination

$$ab = \alpha 1 + \beta e_2 e_3 + \gamma e_3 e_1 + \delta e_1 e_2$$

for scalars $\alpha, \beta, \gamma, \delta$. It turns out that each of these bivectors $e_1 e_2, e_2 e_3, e_3 e_1$ squares to $-1$ under the clifford product, and it is customary to denote these quantities by $k = -e_1 e_2$, $i = -e_2 e_3$ and $j = -e_3 e_1$ so that we have

$$ab = \alpha - \beta i - \gamma j - \delta k$$

In other words, voila--we have just derived quaternions. Quaternions correspond to linear combinations of even-graded elements in the clifford algebra built upon $\mathbb R^3$. Mixed-grade multivectors like these prove very useful in describing reflections and rotations, even in higher-dimensional spaces.

$\endgroup$
  • 1
    $\begingroup$ Thanks, this is a great answer! $\endgroup$ – Mike F Mar 12 '15 at 17:04
4
$\begingroup$

I think that the major merit of exterior algebra is the possibility to define an exterior derivative $d(\omega)$ of an $n$-form and prove that:

$d(d(\omega))=0$

and then prove that if $M$ is a compact smooth orientable $(n+1)$-dimensional manifold with boundary, then we have the generalized Stoke's theorem:

$$ \int_M d(\omega)=\int_{\partial M} \omega $$

That is the most beautiful generalization of the fundamental theorem of calculus.

$\endgroup$
  • 4
    $\begingroup$ But this also doesn't use the whole of the exterior algebra, only $\Lambda^n$, $\Lambda^{n+1}$ and $\Lambda^{n+2}$. $\endgroup$ – Oscar Cunningham Mar 11 '15 at 8:52

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.