Talk:Geometric algebra: Difference between revisions
→A possible source of confusion
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::::::::::::::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)
::::::::::::::After thinking about this, I am going to ask you to change your "It is known" statements in the task description. You need to spell these out in more detail. You cannot expect these details to be known by typical contributors to Rosettacode. You cannot even expect these details to be known by typical mathematicians. Only people who are well versed in the geometri/clifford algebra arcana should be expected to have apriori understanding of those details. --[[User:Rdm|Rdm]] ([[User talk:Rdm|talk]]) 03:31, 20 October 2015 (UTC)
:::::::::::::::The axioms define a so-called geometric product. The fact that this geometric product can be applied to a scalar and something else does not make it a scalar product. The expression ''scalar product'' in math means something specific. IIRC it's a symmetric, bilinear, positive-definite form. The multiplication of a scalar by a vector in a vector space for instance is never called scalar product, as this would be a very unfortunate naming collision. IIRC it's called extern scalar multiplication or something like that.
:::::::::::::::The proof that the geometric product defines a scalar product is not too hard. First you define the inner product of two vectors <math>\mathbf{a}\cdot\mathbf{b} = (\mathbf{a}\cdot\mathbf{b} + \mathbf{a}\cdot\mathbf{b})/2</math>. It's straightforward to see that it is a symmetric and bilinear. What's not so obvious is that it is a form, that is that it returns a scalar. To see it you just notice the equality : <math>\mathbf{a}\cdot\mathbf{b} + \mathbf{a}\cdot\mathbf{b} = \mathbf{a}^2 - \mathbf{a}^2 + \mathbf{a}\cdot\mathbf{b} + \mathbf{a}\cdot\mathbf{b} + \mathbf{b} - \mathbf{b} = (\mathbf{a} + \mathbf{b})^2 -\mathbf{a}^2 - \mathbf{b}^2</math>, and this is a real number because it's a linear combination of real numbers.
:::::::::::::::It's not very complicated a proof, but it's quite irrelevant to the task and putting it in the description would spam it imho. I won't add it unless other people complain.--[[User:Grondilu|Grondilu]] ([[User talk:Grondilu|talk]]) 11:21, 20 October 2015 (UTC)
== "Orthonormal basis" ==
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