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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?

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  • $\begingroup$ I really enjoyed Alan MacDonald's online material about GA. I look forward to snagging a copy of that book someday :) $\endgroup$ – rschwieb Jun 21 '13 at 14:09
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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:

Everything in the vector space $\langle S_i \rangle$ spanned by $S_i$ is said to have grade $i$ (excepting 0 for the time being). Things which are sums of elements of different grades do not have any grade defined for them, and they are said to have mixed grade. Grade only applies to things in these special subspaces. The element 0 lies in all of these sets, so one needs to understand 0 as being an exceptional element. (I'm not exactly sure what most references do concerning 0.)

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."

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    $\begingroup$ A multivector that does not consist of terms of a single grade is not said to be ungraded, but rather that it is of mixed grade, also. $\endgroup$ – Muphrid Jun 21 '13 at 14:16
  • $\begingroup$ @Muphrid That's a good point, it would be good to introduce that. I hope you won't mind if I edit that in. $\endgroup$ – rschwieb Jun 21 '13 at 14:17
  • $\begingroup$ Many thanks for an exceptionally helpful answer! $\endgroup$ – kjo Jun 21 '13 at 19:30
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Please refer to the top of p. 97 and also the previous page. See also Definition 6.7 on p. 98.

Alan Macdonald

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It is pretty much the same meaning as dimension, with the additional specificty of 'dimension of a multivector (or clif)' in the manner defined by Geometric algebra, i.e. scalar = grade 0, vector = grade 1, bivector = 2, etc. But you can just read it as dimension.

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