Quadratic forms and congruence

[imnewhere]

Homework Statement

How many equivalence classes under congruence (as in two quadratic forms - n dimensional vector space over field - being congruent if one can be obtained from the other by a change of coordinates) are there when (i) n=4 and field is complex numbers (ii) n=3 and field is real numbers.

Homework Equations

A quadratic form over C has form (x_1)^2+(x_2)^2+...+(x_r)^2 w.r.t. suitable basis and r=rank of quadratic form

A quadratic form over R has form (x_1)^2+...+(x_t)^2-(x_t+1)^2-...-(x_u)^2 w.r.t. suitable basis and t+u=rank of quadratic form

Basically these: http://img175.imageshack.us/img175/3888/propvm.jpg

The Attempt at a Solution

I can imagine what the equivalence class looks like (how it would hold for reflexivity, symmetry and transitivity) but am struggling to calculate how many there would be for given n and field.

I think you'd only have one equivalence class of quadratic forms if the field is the complex numbers (for a vector space of fixed dimension n). Or 1 + 1 + 1 + 1 + 1 = 5 over the complex numbers and n=4 by summing over ranks as the complex case only depends on the rank of q.

With the reals we'd have to choose signs (using the signature). With n = 3 sum over ranks 0 to 4, i.e. 1 + 2 + 3 + 4 = 10 equivalence classes in total over the reals.

And I don't really know where I'm using the propositions given.

I'm just getting really mixed up so any advice as to what I've done right/wrong would be great! Thanks!

 

[mighty2000]

You are on the right track with your thinking. Let's break down the problem into smaller parts to make it easier to understand.

First, let's consider the case where n=4 and the field is the complex numbers. In this case, we are looking at quadratic forms in a 4-dimensional vector space. Now, as you have correctly stated, the equivalence class of a quadratic form in this case depends only on its rank. This is because any two quadratic forms with the same rank can be obtained from each other by a change of coordinates. So, for each possible rank (0, 1, 2, 3, 4), we will have one equivalence class. Therefore, there will be a total of 5 equivalence classes for n=4 and the field is the complex numbers.

Now, let's consider the case where n=3 and the field is the real numbers. In this case, we are looking at quadratic forms in a 3-dimensional vector space. Now, as you have correctly stated, the equivalence class of a quadratic form in this case depends on its rank as well as the signature. The signature is determined by the number of positive and negative terms in the quadratic form. So, for each possible combination of rank and signature, we will have one equivalence class. Therefore, there will be a total of 10 equivalence classes for n=3 and the field is the real numbers.

As for the propositions given, they are just there to help you understand the structure of quadratic forms and how they relate to each other. For example, Proposition 1 shows that any two quadratic forms with the same rank are equivalent. Proposition 2 shows that if a quadratic form has a certain signature, then all other quadratic forms with the same signature are equivalent. These propositions can help you visualize and understand the different equivalence classes.

I hope this helps clarify things for you. Keep in mind that these are just the basic cases and there may be other factors that can affect the number of equivalence classes. But for the purposes of this problem, this should give you a good understanding.

Best of luck with your research!A fellow scientist