What is the meaning of "identify", as in "$a$ is identified with $b$"?

$\begingroup$

In abstract algebra, there is a terminology called "$a$ is identified with $b$".

But I can't find the formal definition for this meaning. Sometimes we may use the symbol $\bar{a} $ to state that $\bar{a}$ is identified by $a$.

One specific example:

  • A linear function that evaluates a point $a$ is identified with point$a$. (In this case, it's just an isomorphism between $a$ and this function)
  • An adjoint operator $A^{**}$ is identified with $A$. (This case is a bit more complex. That is, for each value that $A^{**}$ map, there is some link of transformation from this value to value that $A$ evaluate.)
  • We can identify action $A$ with action $B$

What is the formal definition for them? I know there is something to do with isomorphism but not so directly?

$\endgroup$ 2

1 Answer

$\begingroup$

Identifying elements of one set with elements of another set means that you have identified an unambiguous isomorphism, so that it doesn't really matter whether you use one or the other. For convenience they are then used interchangeably, as if they were the same. That is, we declare the difference between them as irrelevant. It is understood that if you have an object of the first kind, and need an object of the second kind, you are supposed to just use this special isomorphism.

A simple example is in the classic construction of the integers from pairs of two natural numbers. At some point, it is noted that the equivalence classes of pairs of the form $(n,0)$ behave exactly like the corresponding natural number $n$ (for example, $[(n,0)]+[(m,0)]=[(n+m,0)]$), and therefore we can treat them as if they were the actual natural numbers. And indeed after doing so, we “forget” the whole construction of the integers, and claim that $\mathbb N\subset\mathbb Z$, although from the standard construction, we just get an isomorphism between $\mathbb N$ and a subset of $\mathbb Z$.

Now for example if you are given a natural number $n$, and you write down $-n$, you see that this is not an operation defined on the natural numbers, so strictly speaking you need to use this special isomorphism to “translate” the natural number to an integer number, and then negate that. Of course, we denote the natural number and the corresponding integer with the very same symbol (the same string of digits), so in this case we don't even notice that we formally deal with different objects here.

$\endgroup$

Your Answer

Sign up or log in

Sign up using Google Sign up using Facebook Sign up using Email and Password

Post as a guest

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

You Might Also Like