# A question about isomorphism

1. Sep 19, 2009

### kof9595995

I used to think 1 by 1 matrix is a scalar, but someone argued with me and said they were different. Then I tried to convince him that we actually couldn't find the difference between their fields. He then told me the fields were just isomorphic, so he still didn't agree with my opinion.
I can't say he's wrong, but it's just kind of ridiculous to me. It's like saying that one, two three.......are not 1,2,3....., they are just isomorphic.
Can somebody explain it for me?

Last edited: Sep 19, 2009
2. Sep 19, 2009

### HallsofIvy

Staff Emeritus
Well, what do you think "isomorphic" means? To say that two groups, or rings, or fields are isomorphic means that they are exactly the same, just with a different labeling. Aren't "one, two, three, ..." and "1, 2, 3, ..." exactly the same, just with different labeling? The two of you are saying exactly the same things but in different words. Whether or not there is any difference at all in saying "A and B are the same" or saying "A and B are isomorphic" depends on whether or not you think the labeling is important.

3. Sep 20, 2009

### kof9595995

Thanks, you confirm my opinion.

4. Sep 20, 2009

### Talisman

Well, it also kind of depends on what you're trying to do with your "scalar." If you want to multiply it by another matrix, you're gonna have a hard time (it has to have one row). If it were "really" a scalar, you'd just scalar multiply by that matrix.

If I ask you how many letters are in "one", you'd give a different answer than the same question about "1." In relevant ways, they may be the same, but you can always find ways in which they behave differently.

5. Sep 20, 2009

### kof9595995

I think the multiplication of scalars and matrices is also kind of definition, it's not that "natural", without the definition it's also hard enough to do the multiplications of scalars and matrices.
So if you want I can also define the multiplication of 1 by1 matrix and an arbitrary matrix. So I don't think one should overwhelm the other because of this.
I think to study a mathematical object, we must put in a certain relation, like fields or something. If you say "one" and "1" are different because of the number of letters, I can't say it's not right, but I think they just don't differ “mathematically”

6. Sep 20, 2009

### jambaugh

Technically the product of a 1x1 matrix and say a 2x3 matrix is undefined. It is not impossible to redefine the tensor algebra so that 1x1 matrices and scalars are equal rather than just isomorphic. But one is "redefining" so stepping outside the standard definitions. He is technically correct but one is splitting hairs here.

Remember we don't mine matrices and scalars from veins of rock. They are conceptual inventions so a.) we can decide to redefine things differently, and b.) when asking such questions we must go back to the definitions and axioms.

Note that this also means we may have two different 2x2 matrices with identical entries but not being actually equal mathematical objects. One may express a linear operator while the other a bilinear form. They will transform differently under general changes of basis. Note in this case they are form identical but neither equal nor isomorphic.

Example: The identity operator and the Euclidean metric (in an orthonormal basis). Note the identity operator looks the same in all bases while the Euclidean metric is only form equivalent to the identity matrix in an orthonormal basis.

7. Sep 20, 2009

### Talisman

As another pertinent example: in R^3, obviously the xy and yz planes are isomorphic, but if you're actually _doing_ things in that space, you'll confuse yourself if you think of them as the same plane.

In QM, you'll have operators acting on all sorts of subspaces of Hilbert space, but confusing them would be bad. In particular, the "same" idea helps one make sense of operator commutativity: vaguely, two operators acting on "different" spaces will commute, even if both of those spaces "are" C^2 or something.

8. Sep 22, 2009

### kof9595995

I think I've roughly got the idea. For two isomorphic structures, if you focus on the internal structures of each, you actually can't distinguish them. Like you are confined in x-y plane then suddenly you are moved to x-y plane and still confined, then you'll say you are still in the same plane. When we can distinguish x-y and x-z plane, actually we are using a higher class mathematical structure, in this case the 3-d space.
So despite that different symbols can form an isomorphism, but the difference in symbols probably indicates there exists a higher class of structure (although not necessarily), in which you can distinguish them.

9. Sep 23, 2009

### jambaugh

Sound like you have a good grasp of it. The formal definition of two mathematical object being isomorphic is that there exists an isomorphism mapping between them.

An isomorphism is a mapping which preserves structure relations and is invertible.
It is a special case of a homomorphism which is a structure preserving map (but is not necessarily invertible.)

Examples: Column vectors and row vectors are isomorphic as vectors. The transpose operation gives us an isomorphism mapping.

The multiplicative structure of NxN matrices is homomorphic to the multiplicative structure of real numbers with the determinant mapping since det(AB) = det(A)det(B). Since you can't recover the matrix from just knowing its determinant this is clearly not an invertible mapping and so this homomorphism is not an isomorphism.

Now one must be careful to identify the context (category) in which one is referring to structure. As linear spaces (vectors) 2x2 matrices are isomorphic to 1x4 matrices. But obviously not when you consider the additional multiplicative structure.

Note all vectors spaces (over the same scalar field R, C,...) are isomorphic as linear spaces if they have the same dimension.

Other "morphisms" are endomorphisms which are homomorphisms from a mathematical object into itself. Example projection of vectors onto a subspace is an endomorphism.

When an endomorphism is invertible (over the whole object) it is an automorphism. I.e. an automorphism is both endomorphism and isomorphism. This invertibility makes the set of automorphisms form a group.

You can think of NxN matrices as the endomorphisms on the 1xN column vectors by left multiplication. Linear operators preserve the linear structure of a vector space and so are the endomorphisms of that space. Invertible matrices yield automorphisms and so the automorphism group of an N-dimensional vector space is the group GL(N) of invertible NxN matrices.

When we add more structure to the space like a metric or inner product we get smaller groups O(N) or U(N).

There has been a trend in mathematics to define objects such as groups, fields, rings, vectors spaces, algebras.... in terms of their morphism structure. This subject called category theory.

I know this may be "too much information" but it may help put the concept of isomorphic objects in some context.

10. Sep 24, 2009

### kof9595995

Well. thanks,it's really helpful. I'll take my time to read it.

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