Talk:Geometric algebra: Difference between revisions
→A possible source of confusion
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:::::::::::::No, the axioms do not require that a scalar product can be applied to i, j and k. I think you're confusing things. The axioms imply that the ''geometric product'' can be applied to them, not the scalar product. In fact, the axioms do not mention a scalar product at all. Also I don't understand where you got the idea that I was suggesting that the geometric product is required for the definition of a vector.--[[User:Grondilu|Grondilu]] ([[User talk:Grondilu|talk]]) 20:57, 19 October 2015 (UTC)
:::::::::::::As far as the orthonormality is concerned, once a scalar product has been established, the definition of orthonormality is common knowledge enough that we don't have to remind it, imho.--[[User:Grondilu|Grondilu]] ([[User talk:Grondilu|talk]]) 20:59, 19 October 2015 (UTC)
::::::::::::::Specifically, the axioms include:
:<math>\begin{array}{c}
a(b+c) = ab+ac
\end{array}
</math>
::::::::::::::And as I understand it, a is a scalar or can be a scalar, and i, j and k are values which can be used in the context of b and/or c. Do you really disagree? If so, why?
::::::::::::::That said, scalar product itself has two relevant meanings in this discussion - a product between vectors which produces a scalar, and a product between a scalar and a vector which produces another vector. But even there, it's my understanding that a vector space only needs to support the ability to be scaled and added. Once you have that you have enough that you can easily define a mechanism which multiplies vectors and produces a scalar. That said, you have already defined product involving i, j and k which produces a scalar - that particular product doesn't make them orthonormal, but it would be easy enough to define another product which does. <code>-mul</code>, for example. --[[User:Rdm|Rdm]] ([[User talk:Rdm|talk]]) 02:49, 20 October 2015 (UTC)
== "Orthonormal basis" ==
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