Quadratic forms and congruence

  • Thread starter imnewhere
  • Start date
  • #1
1
0

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 [Broken]

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!
 
Last edited by a moderator:

Answers and Replies

Related Threads on Quadratic forms and congruence

  • Last Post
Replies
0
Views
888
  • Last Post
Replies
3
Views
2K
Replies
4
Views
4K
  • Last Post
Replies
0
Views
2K
  • Last Post
Replies
3
Views
3K
  • Last Post
Replies
1
Views
2K
  • Last Post
Replies
2
Views
2K
Replies
0
Views
1K
  • Last Post
Replies
6
Views
727
  • Last Post
Replies
1
Views
6K
Top