I'm reading a book (Linear and Geometric Algebra, by Alan Macdonald) where the author uses the term grade without ever defining it. I have a murky sense of what the grade of a blade may be (a geometric product of k orthogonal vectors is called a blade of grade k), but I have no idea of what the grade of a general multivector is. Does grade apply only to blades? If not, what is the definition of grade?
[Math] the “grade” in geometric algebra
abstract-algebraclifford-algebrasgeometric-algebras
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There is a definition of the geometric product that applies to general multivectors in any Clifford algebra. It follows directly from the definition of the Clifford algebra. To define a Clifford algebra you need a vector space $V$ and a symmetric bilinear form $B(u,v)$ defined for any $u,v\in V$. The Clifford algebra is the quotient of the tensor algebra of $V$ with respect to the two-sided ideal generated by all elements of the form $u\otimes v+v\otimes u -2B(u,v)$ where $u,v\in V$. The geometric product is the product in the quotient algebra. It is standard and you can find the definition of that in any textbook on abstract algebra. Basically, the geometric product is the product in the tensor algebra of $V$ modulo the ideal.
AN EXAMPLE:
To illustrate, consider $\mathbb R^2$ and the bilinear form defined by $B(e_1,e_1)=1$, $B(e_2,e_2)=1$, $B(e_1,e_2)=0$, where $e_1=(1,0)$ and $e_2=(0,1)$. The two-sided ideal generated by $u\otimes v+v\otimes u -2B(u,v)$ is infinite dimensional as the tensor algebra itself. It contain the following elements among others:
$e_1\otimes e_1-1,\quad e_2\otimes e_2-1,\quad \text{and}\quad e_1\otimes e_2+e_2\otimes e_1$.
This can be used to compute the following products:
$e_1e_1 = e_1\otimes e_1=e_1\otimes e_1 -(e_1\otimes e_1-1)=1$,
$e_2e_2 = e_2\otimes e_2=e_2\otimes e_2 -(e_2\otimes e_2-1)=1$,
$e_1e_2=e_1\otimes e_2= \tfrac{1}{2}(e_1\otimes e_2- e_2\otimes e_1)+\tfrac{1}{2}(e_1\otimes e_2+ e_2\otimes e_1)=\tfrac{1}{2}(e_1\otimes e_2- e_2\otimes e_1)$.
In short, $e_1^2=1$, $e_2^2=1$, and $e_1e_2=-e_2e_1$.
Even though the tensor algebra is infinite-dimensional, the quotient algebra is finite-dimensional. See what happens if you try to get to grade 3. For instance, consider this product
$e_1(e_1\wedge e_2)$ where $e_1\wedge e_2=\tfrac{1}{2}(e_1\otimes e_2- e_2\otimes e_1)$.
It is again a straightforward application of the tensor product modulo the ideal:
$e_1(e_1\wedge e_2)=\tfrac{1}{2}(e_1\otimes e_1\otimes e_2 - e_1 \otimes e_2\otimes e_1)$
but $e_1\otimes e_1\otimes e_2=(e_1\otimes e_1-1)\otimes e_2 +e_2=e_2$ and
$e_1\otimes e_2\otimes e_1=(e_1\otimes e_2+e_2\otimes e_1)\otimes e_1 -e_2\otimes e_1\otimes e_1=$
$=e_2\otimes e_1\otimes e_1=e_2+e_2\otimes(e_1\otimes e_1-1)=e_2$.
So, we get $e_1(e_1\wedge e_2)=\tfrac{1}{2}(e_2+e_2)=e_2$, that is we are back to grade 1. Since the quotient algebra is finite-dimensional, every element can be expressed in term of the basis, which consists of $1, e_1, e_2, e_1e_2$ in the example we are considering. So, every multivector $A$ can be expressed as follows:
$A=s+xe_1+ye_2+pe_1e_2$.
If you have two such multivectors you can compute the product simply by using associativity, distributivity, and the properties we have derived above: $e_1^2=1, e_2^2=1, e_1e_2=-e_2e_1$.
You can easily repeat this exercise for other dimensions and for different bilinear forms.
To come back to the definition of the geometric product, here is how you can understand its significance. In geometry, you are dealing with certain geometric structures. For instance, you might want to find a line passing through two points, or you might want to find a point at the intersection of two lines. These kinds of problems can be dealt with efficiently by applying the exterior structure. You also might want to find, say, a line which passes through a given point and is perpendicular to another line. This kind of problem is related to the orthogonal structure. The tensor product is too general. By using the quotient algebra you are effectively eliminating any part of the tensor product which is not related to exterior or orthogonal structure. What is left has a clear geometric significance. In a way, the geometric product does a lot of work for you behind the curtains, so that you can concentrate on the relevant geometric structures. The expression $uv=u\cdot v+u\wedge v$ is not really the definition of the product. It is just a property that the geometric product of two vectors has.
Alan Macdonald does not use the definition of the geometric product I described above because he does not presume his readers are familiar with the tensor algebra, ideals, or quotients. Instead, he wants to concentrate on applications, geometric properties of the algebra, and on computation. If you are not satisfied with his approach, perhaps you need to read another book. This one
Clifford Algebras and Lie Theory, by Meinrenken
is recent and it uses the same definition I used. There are other equivalent ways to define Clifford algebras. If you are interested, check out these books as well
Quadratic Mappings and Clifford Algebras, by Helmstetter and Micali,
Clifford Algebras: An introduction, by Garling.
Perhaps, after trying to read these books you will appreciate Alan's book more.
Clifford algebra is a well-established part of standard mathematics. It is used in differential geometry and Clifford Analysis, not to mention various applications in physics. No one questions its validity. People who refer to it as Geometric algebra simply want to help promote it in engineering, applied mathematics, and physics. The focus is on applications rather than mathematical rigour. As Alan has pointed out, you don't need to know how the algebra is defined in general in order to use it. You can always compute the product in the basis. It gets tedious to do it by hand as the dimension of the underlying vector space increases, but it can be implemented on a computer quite easily.
Let $b=a_1$ and $C_{r-1}=a_2\wedge a_3\wedge\dots\wedge a_r$. Then, from (1.15) it follows that $$ a\cdot(a_1\wedge a_2\wedge\dots\wedge a_r)=a\cdot a_1 a_2\wedge\dots\wedge a_r $$ $$ -a_1\wedge(a\cdot a_2\wedge\dots\wedge a_r). $$ Next you apply the same (1.15) to the second factor in the second piece above with $b=a_2$ and $C_{r-2}=a_3\wedge a_4\wedge\dots\wedge a_r$ to get $$ =a\cdot a_1 a_2\wedge\dots\wedge a_r-a\cdot a_2 a_1\wedge a_3\dots\wedge a_r+ $$ $$ +a\cdot a_3 a_1\wedge a_2\wedge a_4\wedge\dots\wedge a_r+\textrm{next term}. $$ One of the factors in the next term above has a structure similar to the left hand side in (1.15) and you can iterate the above procedure. In the last step, you have $C_1$ which is a vector, and the inner product with $C_1$ is the usual dot product of vectors.
Here is how you proceed: $$ -a_1\wedge(a\cdot (a_2\wedge\dots\wedge a_r))= $$ $$ =-a_1\wedge[(a\cdot a_2a_3\wedge\dots\wedge a_r)-a_2\wedge(a\cdot(a_3\wedge a_4\dots\wedge a_r))]= $$ $$ =-(a\cdot a_2)a_1\wedge a_3\wedge\dots\wedge a_r+a_1\wedge a_2\wedge(a\cdot(a_3\wedge a_4\dots\wedge a_r)) $$ Observe that the "$\cdot$" operation appears in two forms. On the first of the three lines above it stands for the inner product of vector with multivector, whereas on the first term in the line right below it is just a dot product of two vectors hence a scalar. Scalars can be pulled out of any type of product in geometric algebra and that is how you end up with the first term in the third line. You can deal with the second term in the last line exactly in the same way: you get a term containing a scalar $a\cdot a_3$ which can be pulled out and another term of the form $a_1\wedge a_2\wedge (a_3\cdot(a_4\wedge a_5\dots\wedge a_r))$ where $\cdot$ stands again for vector by multivector as in the left hand side of (1.15) and you repeat the procedure to pull subsequent scalars.
Best Answer
The short answer
No, a general multivector is not graded. Only a portion of the elements of the algebra are assigned grades. Yes blades are among the elements assigned grades. A blade represents a subspace of $V$, and the grade of that blade is the dimension of the subspace the blade represents.
Everything I'm about to say is with regards to an $n$ dimensional quadratic space $V$ with a nondegenerate form.
Hopefully you are familiar with the construction where an orthonormal basis $\{e_i\mid 1\leq i\leq n\}$ for $V$ is used to construct a basis for the Clifford algebra.
The basis for the Clifford algebra is made up from these sets:
$S_0=\{1\}$
$S_1=\{e_i\mid 1\leq i\leq n\}$
$S_2=\{e_ie_j\mid 1\leq i< j\leq n \}$
$S_3=\{e_ie_je_k\mid 1\leq i< j<k\leq n \}$
...
$S_n=\{e_1e_2\cdots e_n\}$
The basis for the Clifford algebra we are talking about is $\cup_{j=1}^n S_j$.
What a grade is:
It isn't hard to show that any product of $k$ vectors (vectors, elements of $V$ lie in $\langle S_1 \rangle$) lies in $\langle S_k\rangle$. It turns out that the product is zero exactly when the set of vectors you are multiplying is linearly dependent. Blades, being products like this, do always have a grade.
What grade roughly means:
For a blade $b_1\cdots b_k$ of $k$ vectors from $V$ made up of linearly independent vectors $b_i$", you get an element of grade $k$. Since this element is usually interpreted as "the subspace of $V$ generated by the $b_i$, you can see that the grade is coinciding with the dimension of this subspace.
When the vectors are not LI, things are different. Had the $b_i$ been linearly dependent, the product $b_1\cdots b_k$ would be zero, but the dimension of the span of the $b_i$ could be anything less than $k$, so you can see why there are problems assigning 0 a fixed grade to make the system of "grade is the same as dimension of span" work universally.
The interesting thing is that multiplication plays well with grades. I think the path most often taken is to just define the grade of 0 to be 0, and then declare "the product of a grade $k$ vector with a grade $j$ vector has grade $j+k$ or grade $0$ according to whether the product is zero or not."