Geometrically independent set

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Demote once generalised, this is taken from an algebraic topology textbook. As such it deals with [ilmath]\mathbb{R}^N[/ilmath] not an arbitrary vector space
  • Can we generalise further to a module rather than a vector space? Probably as this is a stone's throw from Linearly independent set
    • Of course we can, see the equivalent statement


Let [ilmath](V,\mathbb{F})[/ilmath] be a vector space over the field [ilmath]\mathbb{F} [/ilmath] and let [ilmath]\{a_0,\ldots,a_n\}\subseteq V[/ilmath] be a set of points. We say this set of points is "geometrically independent" if[1]:

  • [math]\forall (t_i)_{i\eq 0}^n\subseteq\mathbb{F}\left[\left(\sum_{i\eq 0}^n t_ia_i\eq 0\wedge\sum_{i\eq 0}^n t_i\eq 0\right)\implies \forall i\in\{0,\ldots,n\}\subset\mathbb{N}[t_i\eq 0]\right][/math]

Equivalent statement

  • [ilmath]\{a_0,\ldots,a_n\} [/ilmath] is geometrically independent if and only if the vectors [ilmath]\{a_1-a_0,\ldots,a_n-a_0\} [/ilmath] is a linearly independent set[1]
    • We need not choose [ilmath]a_0[/ilmath] as our point to discard and subtract from everything else, we can choose any member. How to write this though...
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Where's the proof?

See also


  1. 1.0 1.1 Elements of Algebraic Topology - James R. Munkres