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I'm following this intro to geometric algebra and I'm a bit confused about what the difference between these two objects is.

The author states (p. 12)

Let r > 1; then an r-blade or simple r-vector is a product of r orthogonal (thus anticommuting) vectors. A finite sum of r-blades is called an r-vector or homogeneous multivector of grade r.

So by this I can infer that an r-blade is the same as a simple r-vector, but not necessarily an r-vector.

But my problem is then, what is meant by finite sum? If I have a sum of a single r-blade, then that seems to meet the criterion for being an r-vector. So it seems like every r-blade is an r-vector.

dylnan
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    My understanding (supported by the choice of terminology) of the passage you quoted is that every $r$-blade is an $r$-vector but not every $r$-vector is an $r$-blade. – Andreas Blass Aug 09 '21 at 01:04
  • @AndreasBlass I see. What feature further defines an r-blade? – dylnan Aug 09 '21 at 01:08
  • According to the quotation, the further feature that defines an $r$-blade is being a product of $r$ orthogonal vectors. – Andreas Blass Aug 09 '21 at 01:19
  • @AndreasBlass Right, but I don't see why a an arbitrary r-vector can't be written as a product of r orthogonal vectors. Any idea why? – dylnan Aug 09 '21 at 01:27
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    Assuming that "$r$-vector" here means what it usually does, only rather special $r$-vectors can be written as products of vectors. For example, starting with a $4$-dimensional vector space with an orthonormal basis ${a,b,c,d}$, can you express $ab+cd$ as a product of two vectors? – Andreas Blass Aug 09 '21 at 01:30
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    @AndreasBlass I see. So adding two 2-vectors that represent different subspaces gives you another 2-vector that can't be written as a product of vectors. This is an aside, but does this mean that r-vectors that aren't blades don't represent any subspace? – dylnan Aug 09 '21 at 01:37
  • Yes, your understanding is correct, and @AndreasBlass can turn the comment into an answer. – Mark S. Aug 09 '21 at 01:42

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Based on my understanding of GA, $r$-vectors are linear combinations of basis $r$-blades. Basis $r$-blades are products of $r$ linearly independent basis vectors that span an $r$-dimensional subspace embedded in some $d$-dimensional space (where $r \leq d$). For such a $d$-dimensional space, one can construct $\left( \begin{array}{c}d\\r\end{array}\right)$ ($d$ choose $r$) independent $r$-blades. Based on the choice of the basis vectors the constructed $r$-blades will span different oriented sub-spaces. If a specific $r$-vector is not fully embedded in one of these oriented sub-spaces, then it will have to be expressed as a linear combination of more than one $r$-blade. With that being said, another choice of $r$ basis vectors might allow for a single basis $r$-blade to span the specific sub-space containing the desired $r$-vector. This is similar to how one can redefine the set of basis vectors in a 3D space in such a way as to align a single one of those basis vectors (1-blade) with any arbitrary vector (1-vector) direction.

CollinsEM
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