Math 216: Foundations of algebraic geometry 2007-08 of Prof. Ravi Vakil

Here is the materials of Prof. Vakil ‘s course on Foundations of Algebraic Geometry 07-08 in Stanford University:

Classical Lojasiewicz inequalities

Let {f : \mathbb{R}^n \rightarrow \mathbb{R}} be a real analytic function with {f(0) = 0}. Let {V := \{x \in \mathbb{R}^n | f(x) = 0\}} and {K} be a compact subset in {\mathbb{R}^n}. Then the (classical) \L ojasiewicz inequality asserts that:

  • There exist {c > 0, \alpha > 0} such that

    \displaystyle |f(x)| \ge cd(x, V)^\alpha\quad \text{for}\ x \in K.\ \ \ \ \ (1)

\noindent Let {f : \mathbb{R}^n \rightarrow \mathbb{R}} be a real analytic function with {f(0) = 0} and {\nabla f(0) = 0}. The \L ojasiewicz gradient inequality asserts that:

  • There exist {C > 0, \rho \in [0, 1)} and a neighbourhood {U} of {0} such that

    \displaystyle \|\nabla f(x)\| \ge C|f(x)|^\rho\quad \text{for}\ x \in U.\ \ \ \ \ (2)

As a consequence, in (1), the order of zero of an analytic function is finite, and if {f (x)} is close to {0} then {x} is close to the zero set of {f}. However, if {K} is not compact, the latter is not always true and the inequality (1) does not always hold. The inequality (2) is similar to (1), it is not true in the case of K is non-compact.

Two inequalities (1) and (2) have some special cases. For example, in the inequality (1), if {f} has only isolated zero, i.e. {V = f^{-1}(0) = \{(0, 0, \dots, 0)\}}, this implies {d(x, V) = \|x\|}. Hence, we have

\displaystyle |f(x)| \ge C\|x\|^\alpha, \text{for}\ x\in K.

On the other hand, being different from (2), we have another inequality:

\displaystyle \|\nabla f(x)\| \ge c\|x\|^\beta, \text{for}\ x \in U.

There are some relations between {\alpha, \beta} and {\rho} in complex case and real cases…

Curve selection lemma

There are many versions of the curve selection lemma.

In o-minimal structures, we have to consider definable curves, definable functions, deinable sets,… The definition of definable sets,… we can find in many documents.

Let fr(A) be frontier of A, ie fr(A) = \bar{A} - A . We have:

Curve selection lemma: In the o-minimal structure \mathcal{O}. If x \in fr(A), then there is a definable map \gamma: [0, 1) \to \mathbb{R}^n such that \gamma(0,1) \subset A and \gamma(0) = x .

Puiseux series

In this note, we discuss about Puiseux series and its appearance when we solve the equation f(x,y) = 0. We refer to the book “Algebraic curves” of R.Walker.

Puiseux series are fractional power series:

\bar{a}(x)= a_1x^{\frac{m_1}{n_1}}+a_2x^{\frac{m_2}{n_2}}+ \dots where a_i \ne 0, m_1/n_1 < m_2/n_2 < \dots.

Order of series: O(\bar{a}(x)) = m_1/n_1.

Theorem. K(x)^* is algebraically closed.

(K(x)^* – the fieldof fractional power series).

By the proof of this theorem, we can see that f(x,y) = 0 (an algebraic curve), we can solve \bar{y}(x) (Puiseux series) such that f(x, \bar{y}) = 0.


Nullstellensatz in the real case

In alegebraic geometry, we have Hilbert Nullstellensatz and in real algebraic geometry we have Real Nullstelllensatz. There is a difference between these theorems.

Strong Hilbert Nullstellensatz: I(V_{\mathbb{C}}(I)) = \sqrt{I}.

If polynomial f is vanish on the set \begin{cases}f_1 &= 0 \\ &\dots \\ f_k &= 0\end{cases} (in \mathbb{C}^n ) then f has  the following form:

f \in\sqrt{I} , that is: \exists m \in \mathbb{N}: f^m = g_1f_1 + \dots + g_kf_k với g_j \in \mathbb{R}[X_1, \dots,X_n].

In the case of \mathbb{R}^n

Real Nullstellensatz: I(V_{\mathbb{R}}(I)) = \sqrt[\mathbb{R}]{I}.

V_{\mathbb{C}}(I) \cap \mathbb{R}^n = V_{\mathbb{R}}(I):

f^{2s} \in -(\sum \mathbb{R}[X]^2 + I)  or f^{2s} + \sum_{j =1}^m p_j^2 = h_1f_1 + \dots + h_kf_k .

Weak Nullstellensatz

In algebraic geometry, one of  the fundamental theorems is Hilbert’s Nullstellensatz.

Weak version:

Weak Hilbert’s Nullstellensatz: If k is algebraic closed field then the maximal ideals of k[x_1, \dots, x_n] are exactly of the form (x_1 - a_1, \dots, x_n -a_n) for some a_i \in k.

Some cases:

If k = \mathbb{C}, a system of polynomials equation have a root ’cause V(I) \ne \emptyset.

If k = \mathbb{R}, there exists an ideal I \ne (x_1 - a_1, \dots, x_n -a_n) such that V(I) = \emptyset.

Another version:
\begin{theorem} Ideal I \subseteq k[X_1, \dots, X_n] with k is algebraic closed. Then V(I) = \emptyset implies I = k[X_1, \dots,X_n].

Moreover, k = \mathbb{C}, if we have a system of polynomials equation (f_1 = 0, f_2 = 0, \dots, f_k = 0),  if this one have no root then there exist g_1, \dots, g_k \in \mathbb{C}[X_1,\dots,X_n]

s.t. f_1(X)g_1(X) + f_2(X)g_2(X) + \dots + f_k(X)g_k(X) = 1.

For example (Parrilo): Consider following polynomials over \mathbb{C}:

f_1(x) = x^2 + y^2 - 1 =0,

f_2(x) = x + y = 0,

f_3(x) = 2x^3 + y^3 +1 = 0.

There exist:

g_1(x) = \frac{1}{7}(1 - 16x - 12y - 18xy - 6y^2),

g_2(x) = \frac{1}{7}(-7y - x +4y^2 - 16 + 12xy + 2y^3 + 6y^2x),

g_3(x) = \frac{1}{7}(8 + 4y)

s.t. f_1g_1 + f_2g_2 + f_3g_3 = 1.