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I'm reading a book on Clifford algebra for physicists. I don't quite understand it conceptually even if I can do most algebraic manipulations. Can some-one teach me what the Clifford algebra really is? (Keep in mind that I don't know abstract algebra, nothing except some group theory.) Does it make sense to write the sum of a scalar and a bivector in the Clifford product? Both are very different things.

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I cannot make sense of «write the sum of a scalar and a bivector in the Clifford product». –  Mariano Suárez-Alvarez Dec 18 '12 at 15:34
    
The Clifford product of two vectors is the sum of their inner product and the exterior product. –  physicsStudent Dec 18 '12 at 15:35
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The Clifford algebra is a quotient of the tensor algebra. In the tensor algebra you can write the sum of a vector an a scalar as a formal sum. –  Phira Dec 18 '12 at 16:10
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In a certain sense, the Clifford algebra is exactly what you get when you want to identify (certain) products of vectors with scalars. –  Phira Dec 18 '12 at 16:11
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I want to understand not in the most general case . Only the simplest. –  physicsStudent Dec 18 '12 at 16:25

3 Answers 3

The Clifford (or "geometric") algebra is an algebra "generated" by a vector space with a bilinear form with some special properties. There are several pockets of physicists and mathematicians experimenting and refining how to use these algebras to do some geometry.

If you like, you can think of the real 3-D space as your vector space $V$ with its regular inner product, and now imagine it sitting inside an algebra $Cl(V)$. (I am omitting a lot of details in the interest of simplicity.) Since $V$ is three dimensional, so you expect $Cl(V)$ to have at least that many dimensions, but it turns out it will be $2^3$ dimensional (and $2^n$ dimensional for an $n$ dimensional $V$.)

Since $V$ is inside of an algebra $Cl(V)$ there is a multiplication defined between every two elements of $Cl(V)$, and in particular between any two elements $x,y$ of $V$. In fact, things are set up so that $v^2=v\cdot v$, where the right hand side is the inner product in 3-space, and $v\in V$.

Clearly there are many more elements in $Cl(V)$ than just those in $V$. The question is: do they have a useful interpretation? The answer is "yes", because of the way things are set up. It turns out that subspaces of $V$ can be represented as products of elements of $V$ in $Cl(V)$. Because multiplication is defined everywhere, you can now "multiply things in $V$ with subspaces of $V$".

Again, because of how things are set up, you can use this multiplication to do a lot of useful things. For example, you can determine if an element of $V$ is in a certain subspace, or you can reflect a vector in a subspace, or do rotations about a vector in $V$, all using the product of this algebra (and a couple of derived products which I'm not mentioning.)

I'm going to throw in here that when you stop using the usual inner product on $\Bbb{R}^3$, and you switch to the zero bilinear form (that just says "$x\cdot y=0$ for all $x,y\in V$") you get exactly the exterior algebra of $V$ another famous and useful algebra-geometry topic.

That is a very skimpy introduction. I can recommend a lot of resources on the topic, though. If you are very physics oriented, then I think you will enjoy Macdonald's online pdfs on the topic. They are very easy to read. There is also a text (which I have browsed but not really read yet) by Doran and Lasenby called "Geometric algebra for physicists". Another good link, and even an MIT course,another good one: Lundholm-Svensson, good general info.

If you ever want to read it from an algebraic perspective, Clifford algebras appear in Jacobson's Basic Algebra volumes (one or both), and they also appear in Artin's Geometric Algebra. (However, be aware that Artin's title is meant to be taken literally, and it is not really referring to what Doran, Lasenby et al are referring to.)

You will find no shortage of information on "geometric algebra" online, since it is not really a popular visible topic, and the promoters must keep the information freely flowing. I say it's not popular "not very visible" in the pure math community, but it sure is an interesting and fun topic. I imagine as physicists and computer scientists continue to find applications for it, the pure math community will eventually get interested too.


By the way, I saw you noted "$ab=a\cdot b+a\wedge b$" and commented on the exterior product. The wedge there is one of the products that I skipped mentioning for now. It shares many properties with the exterior-algebra-exterior-product.

The identity $vw=v\wedge w+v\cdot w$ works for $v,w\in V$ "because of how things are set up." Surprisingly, the identity also works if you put a few things other than vectors into it, too.


I also thought I would give what I thought were the most elementary cases you might start looking at. Certainly 3-D space with the usual (+++) inner product is where to start. Later, you might get interested in the case where the signature of the inner product is (+---) or (-+++).

Before you do the 4-d space though, you might want to puzzle out how $\Bbb{C}$ and $\Bbb{H}$ are Clifford algebras for a 1 dimensional real vector space with signature (-), and for a two dimensional space with signature (--) respectively.

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It might be misleading to say that Clifford algebras are not "popular" in the mathematics community, given that the Dirac operators that appear in the Atiyah-Singer Index Theorem are Clifford-algebra valued, and so on. True, one will rarely see these things in the standard "required first-year grad curriculum", but that's a different matter. –  paul garrett Dec 18 '12 at 18:44
    
For the physical contexts in which they're of interest, it may be informative to look at complex numbers and quaternions as even subalgebras for the 2d and 3d Euclidean spaces instead. This leads naturally into the notion of spinors and emphasizes how both are unified by clifford algebra in a general formalism for rotation representations. –  Muphrid Dec 18 '12 at 18:52
    
@paulgarrett Yeah, you're right. I just don't have a better word for it. I just mean to express that it is kind of hard to see among "popular" areas like algebraic geometry and differential geometry. –  rschwieb Dec 18 '12 at 18:55
    
Maybe "visibility" is the word...? But if we do emphasize that Clifford algebras are part of the tradition of trying to understand "higher-dimensional numbers" (complex, quaternions), we could easily claim that although the name "Clifford algebra" may not be as visible as some things, the ideas are widely used, whether known by that name or not. –  paul garrett Dec 18 '12 at 19:19
    
@paulgarrett Good alternative! –  rschwieb Dec 18 '12 at 20:01

Let's consider a Euclidean 3d space.

The clifford algebra on this space consists of multivectors, which have 8 linearly independent components. These components can be broken down as follows:

  • 1 scalar component
  • 3 vector components
  • 3 bivector components, which correspond to the 3 linearly independent planes in a 3d space
  • 1 trivector or pseudoscalar component, which corresponds to the single, oriented unit volume in 3d space

(Note the 1,3,3,1 pattern. An N-dimensional space will have $2^n$ components for its multivectors, broken down in this way as according to Pascal's triangle.)

Because all of these components belong to a single multivector object, it does make sense to say you can add scalars and bivectors. You just add corresponding components as you would with traditional vector algebra.

The clifford algebra of 3d space is the algebra of these multivectors--so you have addition and subtraction as usual. Clifford algebra also makes possible various products of multivectors, which lend themselves to simple geometric interpretations.

Beyond that, if you have a more specific question, I'd be happy to try to answer it. As far as saying what clifford algebra is (from the perspective of a lowly physicist), this is all I can say. Why it's useful or how one uses it on a practical level is another question entirely.

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@rschwieb Anything in particular you're looking for input on? Seems like there's been a lot of discussion on what that intro should be. I don't want to step into some points that have already been settled to satisfaction. –  Muphrid May 16 '13 at 16:03
    
I've left a note on your talk page. –  Muphrid May 16 '13 at 16:47
    
Thanks :) ${}{}{}$ –  rschwieb May 16 '13 at 16:55

From a 'what makes it tick' viewpoint, here is a function that generates Clifford Algebras with any signature, in pseudocode:

for nbr_generators = 0 to Whatever \n

max = 2^n - 1
for signature = 0 to max   // all possible signatures +++ to ---...
   for i = 0 to max        // multivectors in LEXICOGRAPHIC order !
      for j = 0 to max 

         k = ( i XOR j )
         sign_ac = anticommuting_sign( i, j )
         sign_sig = signature_sign( i, j, sig )
         net_sign = sign_ac * sign_sig 
         Result[ k ] = A[i] * B[j] * net_sign  // the product
         Cayley_table[i,j] = net_sign 

signature_sign( i, j, sig ) // bit is ON if generator squared = -1

  squares = ( i AND j )
  negative_squares = ( squares AND sig )
  if ( bitcount( negative_squares ) ) is odd
     return -1
  return 1

anticommuting_sign( i, j )

  convert i and j to strings - each ON bit to a letter
  concatenate string_j to string_i
  do a bubble sort counting exchanges of distinct letters
  if nbr_exchanges is odd
     return -1
  return 1
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