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I'm reading Herstein's Topics in Algebra and Halmos's Finite Dimensional Vector Spaces. I think I understand that if $V_F$ is a vector space over $F$ and $W$ a subspace of $V$, then

  1. there is an equivalence relation for $u$ and $v$ in $V$ defined by $u \equiv v \pmod W$ when $(u - v) \in W$
  2. the equivalence classes of the relation correspond to the elements of the quotient space $V/W$.
  3. therefore a quotient space is always a partitioning of the space into equivalence classes (of the corresponding modulus relation).

I haven't seen any reference to the converse, and this is my main question: Does an equivalence relation always have a corresponding quotient space structure ? I suspect not, and I think that the following is a valid counter-example:

Define the relation $\leftrightarrow$ between $u$ and $v$ by $u \leftrightarrow v$ if $u = \lambda v$ for some non-zero $\lambda \in F$. This appears to be an equivalence relation where the $0$ vector is in a class of its own, and "co-linear" vectors are in corresponding equivalence classes. I can't see any way (of course, that isn't a proof) to set this up as a modulus relationship, and therefore conclude that it isn't represented by a quotient space. So, a secondary question is: Is this a valid equivalence relation, and is it representable as a quotient space - if not, what's the proof ?

(Depending on feedback, I may have a further related question about tensor products)

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    $\begingroup$ No time for a proper answer, but your equivalence relation is fine. The quotient space exists, but it's not a vector space: It is instead a projective space (plus an extra point coming from the zero vector). Can it be made into a vector space? Sure, but not in a way that makes a whole lot of sense. $\endgroup$ – Harald Hanche-Olsen Feb 12 '15 at 9:51
  • $\begingroup$ You don't have a quotient-space because the relation $u=\lambda v$ for some $\lambda$ doesn't define a subspace. Note the definition of your relation is not very far from the definition of the projective space associated to $V_F$. $\endgroup$ – Bernard Feb 12 '15 at 10:01
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Note that all equivalence relations you get by looking at equality modulo subspaces of $V$ have the special property that $v \equiv w$ and $v ' \equiv w'$ implies $v + \lambda v' \equiv w + \lambda w' $ for all $\lambda \in F$, that is, they are compatible with the vector space structure on $V$. Such an equivalence relation is also sometimes called a congruence relation. But of course, not all equivalence relations on $V$ are congruence relations and you already found an example for this.

However, if you restrict your focus on congruence relations, then you will get indeed a bijection $$ \{W \:|\: W \text{ a subspace of } V\} \rightarrow \{\equiv \:|\: \equiv \text{ a congruence relation on } V\} $$ by mapping a subspace $W$ to equality modulo $W$. The inverse is given by taking the equivalence class of the zero vector.

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  • $\begingroup$ Thanks. So is it the case that an equivalence relation on a vector space corresponds to a quotient space if and only if it is a congruence relation ? $\endgroup$ – Tom Collinge Feb 12 '15 at 13:13
  • $\begingroup$ Yes, that is correct. $\endgroup$ – Matthias Klupsch Feb 12 '15 at 13:29

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