6
$\begingroup$

Suppose an algebraic variety $V$ is given as the solutions to $q$ polynomial equations of degree $\le k$ with real coefficients $$p_1(x_1,\dots,x_m)=0,\dots,p_q(x_1,\dots,x_m)=0$$ for $x\in\mathbb R^m$. It is a theorem due to Milnor that the sum of Betti numbers of $V$ is bounded by $k(2k-1)^{m-1}$. In particular this also bounds the number of connected components of $V$.

Question: Is this bound on the number of connected components of $V$ the best known bound? I am only interested in the case $k=2$. I suspect that for the varieties I am studying the number of connected components is bounded linearly in $m$.

Milnor, John W., On the Betti numbers of real varieties, Proc. Am. Math. Soc. 15, 275-280 (1964). ZBL0123.38302.

$\endgroup$

1 Answer 1

4
$\begingroup$

No. There are better bounds for particular cases (for instance for $3$ quadrics as given in the reference below), and in general it remains an open problem to find sharp bounds on the maximal number of components of a real affine/projective variety of given degree. The paper here of Degtyarev, Itenberg, and Kharlamov, answers some of the questions above.

Let $B^0_r(N)$, $0\leq r\leq N − 1$, denote the maximal number of connected components that a regular complete intersection of $r + 1$ real quadrics in $\mathbb{P}^N_{\mathbb{R}}$ can have. Then, it is straightforward to show that the number of components can grow exponentially in the dimension for a suitable number of quadrics

$B^0_{N−1}(N) = 2^N$ for all $N > 1$ (intersection of dimension zero).

However, if the number of equations is small, for instance three, then there is a quadratic upper bound in $N$ for $B^0_{k}(N)$.

Theorem. For all $N > 4$, one has $\frac{1}{4}(N − 1)(N + 5) − 2 < B^0_2(N)\leq \frac{3}{2}k(k − 1) + 2$, where $k = [\frac{N}{2}] + 1$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.