What is meant by can be identified with

[jpcjr]

What is meant by "can be identified with"

Background

I was reading Anthony Henderson’s paper “Bases For Certain Cohomology Representations Of The Symmetric Group “ (Ref.: arxiv.org/pdf/math/0508162) and came across the following statement in Proposition 2.6 on Page 9:

“V(1, n) can be identified with the subspace of V(r, n) spanned by [T] for T ∈ T (1, n)”.​

While googling the internet trying to understand what is meant by the phrase “can be identified with,” I came across the following statement using the phrase “can be identified with”.

“W* [i.e., the dual space of W] can be identified with the subspace of V* [i.e., the dual space of V] consisting of all linear functionals that are zero.”​

Here is a statement I created in which I believe I used the phrase “can be identified with” correctly:

“Vector space A, a subspace of vector space V, can be identified with vector space B, some other subspace of V.”​

Here’s my question:

Does the phrase “can be identified with” mean the former subspace of V is a subset of and therefore a subspace of the latter subspace of V?

In other words, is the following statement equivalent to the statement above that I created?

“Vector space A, a subspace of vector space V, is also a subspace of vector space B, some other subspace of V.”​

Thanks!

 

[Office_Shredder]

Can be identified with typically means isomorphic. So subspace A of V can be identified with subspace B of V would mean that A and B are isomorphic subspaces. This is not a particularly interesting condition, it just means that they have the same dimension, so the phrase is sometimes used to convey to the reader that the isomorphism is canonical/natural/pick your favorite word that isn't well defined in general usage but which will cause somebody to spit category theory until the thread is locked.

If you aren't familiar with that phrase, it simply means that the isomorphism does not rely on picking a basis. For example if I said "define an isomorphism between {functions of the form a e^x+be^-x} and R²" you might say "a e^x+be^-x gets mapped to (a,b)". But this inherently relied on the fact that our set was being spanned by e^x and e^-x. If I said instead "define an isomorphism between the{functions of the form acosh(x)+bsinh(x)} and R²", you would probably think the previous map is an odd one to use, even though it still works (note the set of functions hasn't changed, just the basis I use to represent them)

On the other hand consider the following statement: V and V^** are isomorphic if V is a finite dimensional vector space. The isomorphism between these (which I won't prove but isn't terribly hard to check) is as follows: given a vector v in V, we define f_v(x) for x in V^* as f_v(x) = x(v). f_v is a linear function from V^* to R, so is in V^**, and this function $v\mapsto f_v$ is actually an isomorphism. This definition doesn't rely on any choice of a basis of V or V^**, so is often referred to as a natural isomorphism or canonical isomorphism, and when people say "can be identified with" they sometimes (but not always) mean the isomorphism is canonical

 

[jpcjr]

Thank you!

By the skin of my teeth, some help from you, and the grace of God, I received the best grade I could have expected in Linear Algebra.

Thanks, again!

Joe