In my number theory class, we're discussing classifications of quadratic forms over various fields. For our homework, we're asked to classify quadratic forms over the complex numbers. From what I understand, over $\Bbb R^n$, we have that every quadratic can be linearly transformed from $$a_{11}x_1^2 + a_{12}x_1x_2 + ... + a_{nn}x_n^2$$ to $$a_1'x_1^2+a_2'x_2^2 + ... + a_n'x_n^2$$ and then to $$x_1'^2+x_2'^2+...+x_p'^2-x_{p+1}'^2-x_{p+2}'^2-...-x_r'^2$$ Where two quadratic forms are equivalent <=> they have the same signature; i.e., the same number for p and r. That last transformation gives us the most general form we can have for quadratics on $\Bbb R^n$, since if we have $ax^2$ for $a>0$, we can transform that using $x=\frac{1} {\sqrt{a}}x'$ so that $x'^2=ax^2$, but if we have $-ax^2$, there is no real root of -1, so we have $-x'^2=-ax^2$ instead of being able to simplify further to $x'^2=-ax^2$. This is why we can't just have every quadratic form over the reals be equivalent to $$x_1^2+x_2^2+...+x_n^2$$ $\mathbf {However}$, over $\Bbb C$, $-1$ has a root, so we should just have that every quatratic form is equivalent to $x_1^2+x_2^2+...+x_n^2$ for some n, meaning two quadratics are equivalent if they have the same rank, right? In fact, this should be the case for any algebraically closed field, correct?

(Please let me know if this has all been reasoned correctly, including the part about quadratic forms over the reals; I want to make sure I understood that argument correctly as well)

  • $\begingroup$ Yes, that's correct. $\endgroup$ – ziggurism Mar 6 '18 at 22:47

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