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## Main Question or Discussion Point

Let the (multi-vector valued) “inner product” between a j-vector U and a k-vector B be defined as the (k-j) grade part of the geometric product UB, (a.k.a. “left contraction”) that is,

$$U\cdot B := <UB>_{k-j}$$

(0 when j > k) as is done in Alan Macdonald’s book “Linear and Geometric Algebra.”

In chapter 7, Alan defines the projection of one blade U in another blade B and then gives a theorem stating it is equal to

$$P_B(U)=(U\cdot B)B^{-1}$$

A natural way to define the rejection of U in B might be as the identity minus the projection, that is,

$$R_B(U):=U-P_B(U)=U-(U\cdot B)B^{-1}$$

This agrees with the previously given (in the book) formula for the rejection of a vector u in a blade B,

##R_B(u)=u-(u\cdot B)B^{-1}=uBB^{-1}-(u\cdot B)B^{-1}=(uB-u\cdot B)B^{-1}=(u∧B)B^{-1}##

This definition of rejection also seems to have the property that it always has zero projection onto B (as we would expect of a rejection) as we can check:

##P_B(R_B(U))=((U-(U\cdot B)B^{-1})\cdot B)B^{-1}=<(U-<UB>_{k-j}B^{-1})B>_{k-j}B^{-1}=<UB-<UB>_{k-j}>_{k-j}B^{-1}=(<UB>_{k-j}-<<UB>_{k-j}>_{k-j})B^{-1}=(<UB>_{k-j}-<UB>_{k-j})B^{-1}=0##

My question is about a comment made by the author in one of the problems. (I can add a picture of the page with the problem if someone confirms it’s allowed.)

He said that this definition of rejection fails in dimensions higher than three. He did not explicitly say the following, but he seems to imply it fails because the projection of the rejection is not always zero... but the above steps seem to at least outline a proof that it is always zero.

So why did he say that this definition is not suitable in higher dimensions??? Have I made a mistake in the above manipulations?

$$U\cdot B := <UB>_{k-j}$$

(0 when j > k) as is done in Alan Macdonald’s book “Linear and Geometric Algebra.”

In chapter 7, Alan defines the projection of one blade U in another blade B and then gives a theorem stating it is equal to

$$P_B(U)=(U\cdot B)B^{-1}$$

A natural way to define the rejection of U in B might be as the identity minus the projection, that is,

$$R_B(U):=U-P_B(U)=U-(U\cdot B)B^{-1}$$

This agrees with the previously given (in the book) formula for the rejection of a vector u in a blade B,

##R_B(u)=u-(u\cdot B)B^{-1}=uBB^{-1}-(u\cdot B)B^{-1}=(uB-u\cdot B)B^{-1}=(u∧B)B^{-1}##

This definition of rejection also seems to have the property that it always has zero projection onto B (as we would expect of a rejection) as we can check:

##P_B(R_B(U))=((U-(U\cdot B)B^{-1})\cdot B)B^{-1}=<(U-<UB>_{k-j}B^{-1})B>_{k-j}B^{-1}=<UB-<UB>_{k-j}>_{k-j}B^{-1}=(<UB>_{k-j}-<<UB>_{k-j}>_{k-j})B^{-1}=(<UB>_{k-j}-<UB>_{k-j})B^{-1}=0##

My question is about a comment made by the author in one of the problems. (I can add a picture of the page with the problem if someone confirms it’s allowed.)

He said that this definition of rejection fails in dimensions higher than three. He did not explicitly say the following, but he seems to imply it fails because the projection of the rejection is not always zero... but the above steps seem to at least outline a proof that it is always zero.

So why did he say that this definition is not suitable in higher dimensions??? Have I made a mistake in the above manipulations?