The Grade Projection Operator $<\cdot>_r$ is widely used in Geometric Algebra to prove numerous relations and results. I'm looking for a proof that Grade Projection is a well-defined operation.
To do as such it should be sufficient to show that an $r$-blade $f_1f_2…f_r$ (geometric product of $r$ normal vectors) becomes an $r$-vector when each $f_{i}$ is expressed in terms of a basis $\{e_1,e_2,…,e_n\}$, meaning:
$$f_1f_2…f_r = \sum_{i,j,…,k} (a_{1i}a_{2j}…a_{rk})e_ie_j…e_k = \sum_{i \neq j \neq … \neq k} (a_{1i}a_{2j}…a_{rk})e_ie_j…e_k$$
Unfortunately, most texts I've found do not bother proving this fact, as they present Geometric Algebra axiomatically. Due to the fact that it remains to be proven that such an axiomatic definition produces an actually existing entity, a constructive approach (be it as a quotient of the tensor algebra or through direct construction from a canonical orthonormal basis) seems to still be necessary.
In An elementary construction of the geometric algebra, Alan Macdonald provides such a construction, including a short "proof" of what I'm looking for. The proof seems to me more of an example than an actual demonstration and moreover, it relies on the principle that you can transform a certain orthonormal basis to another one through a series of reflections operations.
This fact seems to me to be reliant on the fact Macdonald considers a vector space with a positive definite inner product (meaning $a\cdot b \ge 0$ or Euclidean space with Euclidean norm). Geometric Algebra is also used with the Minkowski metric $\{1,-1,-1,-1\}$ for example, for which I do not know if the "consecutive reflections method" for basis vectors transformation still holds.
I'm then looking for a general proof for a Geometric Algebra with signature $\{n,m\}$ (therefore, for any non-degenerate inner product).
Summary:
Prove that the $r$-blade $f_1f_2…f_r$ is expressed as an $r$-vector in another normal basis $\{e_1,e_2,…,e_n\}$ (all components of grade $\neq r$ are null)
N.B. I'm working with the definition of $r$-blade as geometric product of $r$ normal vectors, NOT with the definition using the wedge product. The definition of the wedge product should if possible come AFTER the verification that the Grade Projection Operator is well-defined.
EDIT 1:
I'd like to thank @Nullius in Verba for their answer. I report their answer with expanded algebraic passages (hoping no error has been made and eventually asking for corrections). This is mostly to help myself visualize better their explanation, but I hope this will be of help to others too. The reasoning is/should be the following:
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Consider $r$-blade $f_1f_2…f_j…f_k…f_r$
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Express it in another basis as:
$$f_1f_2…f_j…f_k…f_r=\sum_{i_1,i_2,…,i_j,..,i_k,…,i_r} (a_{i_1}a_{i_2}…a_{i_j}…a_{i_k}…a_{i_r})(e_{i_1}e_{i_2}…e_{i_j}…e_{i_k}…e_{i_r})$$
- Due to the anti-symmetry property of the product of normal vectors (the property can be shown to hold even if working with the $f_i$ basis vectors already expressed in the $e_i$ basis, as shown in An elementary construction of the geometric algebra by Alan Macdonald):
$$f_1f_2…f_j…f_k…f_r+f_1f_2…f_k…f_j…f_r=0$$
- By expressing both in the $e_i$ basis:
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$f_1f_2…f_j…f_k…f_r=
\sum_{i_1 \neq … \neq i_j \neq i_k \neq … \neq i_r} (a_{i_1}…a_{i_j}…a_{i_k}…a_{i_r})(e_{i_1}…e_{i_j}…e_{i_k}…e_{i_r}) +\\
\sum_{\exists i_j,i_k:i_j=i_k} (a_{i_1}…a_{i_j}…a_{i_k}…a_{i_r})(e_{i_1}…e_{i_j}…e_{i_k}…e_{i_r})$ -
$f_1f_2…f_k…f_j…f_r=\sum_{i_1 \neq … \neq i_k \neq i_j \neq … \neq i_r} (a_{i_1}…a_{i_k}…a_{i_j}…a_{i_r})(e_{i_1}…e_{i_k}…e_{i_j}…e_{i_r}) +\\
\sum_{\exists i_j,i_k:i_j=i_k} (a_{i_1}…a_{i_k}…a_{i_j}…a_{i_r})(e_{i_1}…e_{i_j}…e_{i_k}…e_{i_r})$
- The second expression can be rewritten as:
$$f_1f_2…f_k…f_j…f_r=-\sum_{i_1 \neq … \neq i_j \neq i_k \neq … \neq i_r} (a_{i_1}…a_{i_j}…a_{i_k}…a_{i_r})(e_{i_1}…e_{i_j}…e_{i_k}…e_{i_r}) +\\
\sum_{\exists i_j,i_k:i_j=i_k} (a_{i_1}…a_{i_j}…a_{i_k}…a_{i_r})(e_{i_1}…e_{i_j}…e_{i_k}…e_{i_r})$$
- First summation: scalars reordered via commutativity and $e_{i_j}$ and $e_{i_k}$ reordered with the addition of a minus sign in front due to antisymmetry (in the first summation, the condition $i_1 \neq … \neq i_j \neq i_k \neq … \neq i_r$ holds)
- Second summation: indices re-labelled, due to the second summary containing ONLY and ALL members with at least one pair $e_{i_j}$ and $e_{i_k}$ for which $i_j=i_k$ holds (guaranteeing the possibility of re-labelling elements accordingly)
- By summing the two expressions, the following is obtained:
$$f_1f_2…f_j…f_k…f_r+f_1f_2…f_k…f_j…f_r=\\
2\sum_{\exists i_j,i_k:i_j=i_k} (a_{i_1}…a_{i_j}…a_{i_k}…a_{i_r})(e_{i_1}…e_{i_j}…e_{i_k}…e_{i_r})=0$$
Then $\sum_{\exists i_j,i_k:i_j=i_k} (a_{i_1}…a_{i_j}…a_{i_k}…a_{i_r})(e_{i_1}…e_{i_j}…e_{i_k}…e_{i_r})=0$, proving that even if the basis is changed, the only elements which have the possibility of "surviving" the geometric product are precisely those of grade $r$.
Proven this, the Grade Projection Operator $<\cdot>_r$ can be said to be well-defined, as no strange phenomena (appearance of terms of different grades) should happen even if the basis is changed.
N.B. If there exists also a "coordinate-free proof" (which seems to be the motto of Geometric Algebra), which does not require expansion in an orthonormal basis but which follows directly from the properties of Geometric Algebra defined as the quotient of the corresponding tensor algebra, it would be welcome. Nevertheless, I believe that the provided answer is more than satisfactory.
Best Answer
"it relies on the principle that you can transform a certain orthonormal basis to another one through a series of reflections operations"
That's the Cartan-Dieudonné Theorem: For an n-dimensional, non-degenerate symmetric bilinear space over a field with characteristic not equal to 2, every element of the orthogonal group is a composition of at most n reflections. It works fine on the Minkowski metric.
"N.B. I'm working with the definition of r-blade as geometric product of r normal vectors, NOT with the definition using the wedge product."
The concept of 'normal' vectors is probably best defined after defining the inner product, at the same time as the wedge product. For the axiomatic approach, you want to start with algebraic properties of the geometric product and derive the geometric interpretation of thoe properties later. Define r-blades by specifying the vector basis to be pairwise antisymmetric: $f_if_j=-f_jf_i$ for $i\neq j$.
Hopefully with the full Cartan-Dieudonné Theorem in hand, the McDonald proof may be deemed more acceptable. But I don't think it's too hard to prove using the antisymmetry property. Expand the product of $f_i$s writing each as a linear combination of $e_i$s.
$$f_1f_2\ldots f_r=(a_{11}e_1+a_{12}e_2+\ldots +a_{1n}e_n)\ldots (a_{r1}e_1+a_{r2}e_2+\ldots +a_{rn}e_n)$$
Multiply out all the $r$ brackets to get terms with $r$ vectors $e_{i_1}e_{i_2}\ldots e_{i_r}$ in each. If the vectors in a term are all distinct, then the product is an r-blade by the antisymmetry of the $e_i$. If any are repeated, then we can permute the order (with appropriate sign-changes because the $e_i$ are antisymmetric) to put them next to each other and collapse them to a scalar, which commutes under multiplication with any multivector. Now we use this and the antisymmetry of the $f_i$ to show that these must sum to zero. Swapping any pair of the brackets above swaps the order of the corresponding $e_{i_j}$, $e_{i_k}$ in the products, and by antisymmetry of the $f_i$ this must reverse the sign of the sum of all such terms. But if $i_j=i_k$ then $e_{i_j}e_{i_k}=e_{i_k}e_{i_j}$ for all such terms, and the only way this can reverse sign is if they sum to zero.
I am inclined to agree that elementary introductions to Geometric Algebra are lacking. Those directed at physicists do usually try to motivate them geometrically at the start, but at some point the leap to abstraction is made, and from that point on only the most minimal sketch is provided - a scaffold on which, with a lot of work, a full derivation could be constructed. Usually this seems to be because authors want to demonstrate the utility of Geometric Algebra to other professionals by getting on to the more interesting applications quickly. Teaching vector algebra ramps up gradually, from drawing arrows on graph paper at the age of 10 to maybe starting vector calculus (grad, div, curl and all that) at the age of 20. Introductory texts on Geometric Algebra do a couple of pages of drawing arrows to show willing, but only about 10 pages later are discussing Dirac spinors and general relativity and differential geometry! I exaggerate, but not by much. There are about 10 years worth of intermediate detail missing. But such student-oriented texts are unlikely to be written until Geometric Algebra is widely enough accepted and established to be taught at the same elementary stage as vectors are. (Hestenes made the attempt, but again, in my view, tried to do too much at once.) The pedagogy is still a work in progress. I expect it will be another 20 years or more before we get that far.