phaedo Posted August 12, 2015 Share Posted August 12, 2015 Hello, My knowledge of abstract algebra beyond linear vector spaces is very limited. My problem is inspired from the (not very well-known) triangular inequality between angles of a tetrahedron (see for example http://convexoptimization.com/wikimization/index.php/Fifth_Property_of_the_Euclidean_Metric): [latex] \left| \widehat{x,y}-\widehat{y,z} \right| \leq \widehat{x,z}\leq\widehat{x,y}+\widehat{y,z} [/latex] where x,y,z are three vectors and [latex]\widehat{x,y}[/latex] is the angle formed by the vectors x,y. I am looking for a way to abstract this into some algebra of vertices. Say a vertex is the tuple [x,y], we would need to define some "addition" operator in a transitive way so that [x,y]+[y,z] = [x,z], and we would define a norm as [latex]\| {[x,y]} \| = \widehat{x,y}[/latex]. The triangular inequality above would then read in the familiar way: [latex] \left| \| [x,y] \| - \| [y,z] \| \right| \leq \| [x,y] + [y,z] \| \leq \| [x,y] \| + \| [y,z] \| [/latex] Does this look familiar to anyone? Thanks in advance for any pointers. p. 1 Link to comment Share on other sites More sharing options...
ajb Posted August 13, 2015 Share Posted August 13, 2015 [x,y] +[y,z] = [x,z] reminds me of the partial multiplication found in groupoids. You want [x,y]+ [y', z] = [x,z] only if y = y' and leave it undefined otherwise? Link to comment Share on other sites More sharing options...
phaedo Posted August 13, 2015 Author Share Posted August 13, 2015 [x,y] +[y,z] = [x,z] reminds me of the partial multiplication found in groupoids. You want [x,y]+ [y', z] = [x,z] only if y = y' and leave it undefined otherwise? Thanks for pointing groupoids out. It would be nice to find a generalization of addition for any [x, y] + [z, u], but it's not required. One difficulty is that the norm I am considering here is bounded... Not sure if we can make sense of this. Link to comment Share on other sites More sharing options...
ajb Posted August 14, 2015 Share Posted August 14, 2015 One difficulty is that the norm I am considering here is bounded... Not sure if we can make sense of this. Why is having a bounded norm a problem? I do not see that being bounded or not is part of the definition of a norm. You should check that the norm you define really is a norm, that is satisfies the properties you need it to satisfy. i) Absolute homogeneity ii) The triangle inequality iii) Separates points (norm of the zero vector is zero) Link to comment Share on other sites More sharing options...
phaedo Posted August 14, 2015 Author Share Posted August 14, 2015 If norm(a.x) = | a | norm(x) then as | a | goes to infinity the norm of a.x goes to infinity... Thus a norm can't be bounded? Link to comment Share on other sites More sharing options...
ajb Posted August 14, 2015 Share Posted August 14, 2015 Okay, the absolute homogeneity rules bounded out. Good point. Link to comment Share on other sites More sharing options...
Keen Posted November 5, 2015 Share Posted November 5, 2015 What exactly is an algebra of vertices to you? It seems to me that what you are looking for is simply an affine space. Link to comment Share on other sites More sharing options...
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