Lecture Notes on Polynomial Parametrization - PDF

```markdown ## (Lecture 6) ### 44 * We start with a **polynomial parametrization**: $x_1 = f_1(t_1,..., t_m)$ $\vdots$ $x_n = f_n(t_1,..., t_m)$ where $f_1,..., f_n \in \mathbb{K}[t_1,..., t_m]$. $p_0 = (1-t)^2$ $p_1 = 2t(1-t)$ $p_2 = t^2$ * We can think of th...

```markdown ## (Lecture 6) ### 44 * We start with a **polynomial parametrization**: $x_1 = f_1(t_1,..., t_m)$ $\vdots$ $x_n = f_n(t_1,..., t_m)$ where $f_1,..., f_n \in \mathbb{K}[t_1,..., t_m]$. $p_0 = (1-t)^2$ $p_1 = 2t(1-t)$ $p_2 = t^2$ * We can think of this geometrically as the function $\varphi: \mathbb{K}^m \rightarrow \mathbb{K}^n$ $\rho:(t_1,..., t_m) \mapsto (f_1(t_1,..., t_m),..., f_n(t_1,..., t_m))$. * So $\varphi(\mathbb{K}^m) \subset \mathbb{K}^n$ is the subset of $\mathbb{K}^n$ parametrized by the equations. * **NOTE** $\varphi(\mathbb{K}^m)$ need not be an affine variety! So a solution to the implicitization problem concretely means finding the smallest affine variety containing $\varphi(\mathbb{K}^m)$. * **Implicitization Problem (formally)** Given a parametrization as in $f_1,..., F_s$ such that. find $V(F_1,..., F_s)$ is the smallest affine variety containing $\varphi(\mathbb{K}^m)$. * Note that the equations define a variety $V = V(x_1 - f_1,..., x_n - f_n) \subset \mathbb{K}^{n+m}$ * So we can relate implicitization to elimination as follows: ### 45 * Points in V can be written as $(t_1,..., t_m, f_1(t_1,..., t_m),..., f_n(t_1,..., t_m))$ $\Rightarrow V \text{ is the graph of } \varphi$ * **Example**: $\varphi: \mathbb{R} \rightarrow \mathbb{R}$ $t \mapsto t^2 = x$ $(t, x) \in V(\varphi - x)$ * Also have functions: $\iota: \mathbb{K}^m \rightarrow \mathbb{K}^{n+m} $ $\iota: (t_1,..., t_m) \mapsto (t_1,..., t_m, f_1(t_1,..., t_m),..., f_n(t_1,..., t_m))$ * and $\pi_m: \mathbb{K}^{n+m} \rightarrow \mathbb{K}^n$ $\pi_m: (t_1,..., t_m, x_1,..., x_n) \mapsto (x_1,..., x_n)$. * Have the diagram $\mathbb{K}^m \overset{\iota}{\longrightarrow} \mathbb{K}^{n+m}$ $\qquad \searrow \swarrow \pi_m$ $\qquad \qquad \mathbb{K}^n$ * So $\varphi = \pi_m \iota$ * **Exercise** $\iota(\mathbb{K}^m) = V$. $\Rightarrow \varphi(\mathbb{K}^m) = \pi_m (\iota (\mathbb{K}^m)) = \pi_m (V)$ i.e. the image of the parametrization equals the projection of its graph!. $\Rightarrow$ can use the elimination and closure theorems to solve the implicitization problem! ### 46 **Theorem (Polynomial Implicitization)** Let $\mathbb{K}$ be an infinite field, $\varphi: \mathbb{K}^m \rightarrow \mathbb{K}^n$ a polynomial parametrization as in , and define the ideal $I = \langle x_1 - f_1, ..., x_n - f_n \rangle \subset \mathbb{K}[t_1, ..., t_m, x_1, ..., x_n]$. Then $V(I_m)$ is the Zariski closure of $\varphi(\mathbb{K}^m)$ in $\mathbb{K}^n$. **Proof:** We have seen that $\overline{\varphi(\mathbb{K}^m)} = V(I(\varphi(\mathbb{K}^m)))$ is the smallest affine variety containing $\varphi(\mathbb{K}^m)$. * So we want to show that the smallest affine variety containing $\varphi(\mathbb{K}^m)$ is $V(I_m)$. * Start with $V = V(I) \subset \mathbb{K}^{n+m}$ - the graph of $\varphi$. * Assume first that $\mathbb{Z} = \mathbb{C}$. * By we have $\varphi(\mathbb{K}^m) = \pi_m (V)$. * By Closure Theorem, we have $V(I_m)$ is the smallest affine variety containing $\varphi(\mathbb{K}^m)$. - done we $\mathbb{Z} = \mathbb{C}$ * Let $\mathbb{K}$ be a subfield of $\mathbb{C}$. (i.e. a subset of $\mathbb{C}$ that forms a field under the same multiplication & addition). $\Rightarrow \mathbb{K}$ contains $\mathbb{Z}$ (Exercise!) ... and $\mathbb{Q}$! (Why?) $\Rightarrow \mathbb{K}$ infinite. * $\mathbb{K}$ could be non-algebraically closed... can't use closure theorem directly... $V_{\mathbb{Z}}(I_m)$ is variety we get in $\mathbb{K}^n$. $V_{\mathbb{C}}(I_m)$ is variety we get in $\mathbb{C}^n$ $\rightarrow$ Going to larger field doesn't change elimination ideal $I_m$ since the algorithm for computing $I_m$ is not affected by changing from $\mathbb{Z}$ to $\mathbb{C}$. ### 47 * Need to prove $V_{\mathbb{Z}}(I_m)$ is smallest variety in $\mathbb{K}^n$ containing $\varphi(\mathbb{Z}^m)$. *. By and our Lemma, we have $\varphi(\mathbb{K}^m) = \pi_m (V(\mathbb{K}) \subset V_{\mathbb{K}}(I_m)$. Where $V_{\mathbb{K}} = V(x_i - f_1, ..., x_n - f_n)$ (over $\mathbb{K}^{n+m})$ Let $Z_k = V_k (g_1,..., g_s) \subset \mathbb{K}^n$ be a variety in $\mathbb{K}^n$ such that $\varphi(\mathbb{K}^m) \subset Z_k$. * Since $g_i (a) = 0 \; \forall a \in Z_k$ then $g_i (a) = 0 \; \forall a \in \varphi(\mathbb{K}^m)$ $\Rightarrow g_i \circ \varphi$ vanishes on all of $\varphi(\mathbb{K}^m)$ and $g_i \circ \varphi \in \mathbb{Z}[t_1,..., t_m]$. $\Rightarrow g_i \circ \varphi (a) = 0 \; \forall a \in \mathbb{K}^m$ * This mcans $g_i \circ \varphi$ vanishes on all of $\mathbb{K}^m$. Since $\mathbb{K}$ infinite $\Rightarrow g_i \circ \varphi$ is 0-polynomial $\forall i$. $\Rightarrow Z_G = V_k (g_1,..., g_s)$ a variety in $\mathbb{K}^n$ containing $\varphi (\mathbb{K}^m)$. $\Rightarrow V_{\mathbb{C}} (I_m) \subset Z_{\mathbb{C}} \subset \mathbb{C}^n$ (by case $\mathbb{Z} = \mathbb{C}$). * By- considering only solutions in $\mathbb{K}^n$ it follows $V_{\mathbb{Z}} (I_m) \subset Z_{\mathbb{K}}$. proving the result for $\mathbb{Z} \subset \mathbb{C}$. * If $\mathbb{Z}$ not a subfield of $\mathbb{C}$, one can prove there is an algebraically closed field containing $\mathbb{Z}$, and use the same arguments. ### 48 * We get the following algorithm for solving the proliferation problem for polynomial parametrizations: * **Algorithm**: Parametrization $x_i = f_i (t_1, ..., t_m)$ for $i = 1, ..., n$ and $f_1, ..., f_n \in \mathbb{K}[t_1, ..., t_m]$. Define $I = <x_i - f_1, ..., x_n - f_n> \in \mathbb{K}[t_1, ..., t_m, x_1, ..., x_n]$. * Compute Grobner basis $G$ for $I$ with. lexicographic term order. where every $t_j$ is greater than every $x_i$ * Take $G \cap \mathbb{Z}[x_{i}, ..., x_{n}]$ * **Example** $\varphi : \mathbb{R} \rightarrow \mathbb{R}^3$. $\varphi : t \mapsto ((1-t)^2, 2t(1-t), t^2)$ * Compute $I = <p_0 - (1-t)^2, p_1 - 2t(1-t), p_2 - t^2>$ and G.B. * $G = \{p_1^2 + 4p_0p_2 + 4p_2^2 - 4p_2, p_0 + p_1 + p_2 - 1, 2t- P-22\}$. * $ \overline{\varphi(\mathbb{R})} = V (p_1^2 + 4p_0p_2 + 4p_2^2 - 4p_2, p_0 + p_1 + p_2 - 1)$ * $<p_1^2 + 4p_0p_2 + 4p_2^2 - 4p_2, p_0 + p_1 + p_2 - 1>=<p_1^2 - 4p_0p_2 + 4p_2^2 , p_0 + p_1 + p_2 - 1>.$ $\overline{\varphi(\mathbb{R})} = V (p_1^2 - 4p_0p_2, p_0 + p_1 + p_2 - 1)$. $\Rightarrow$ $\overline{M} = \overline{\varphi((0,1)} = \varphi(\mathbb{R}) = V (p_1^2 - 4p_0p_2 , p_0 + p_1 + p_2 - 1)$. ((skipping details...)) $\Rightarrow$ Semialgebraic description of the binomial model for $n = 2$ is $M = V (p_1^2 - 4p_0p_2) \cap \Delta_0^2$ ### 49 **Example** (Concentration Models). $\Xi = [\Xi_1, \Xi_2, \Xi_3]^T \sim N(0, \Sigma)$ $\Sigma^{-1} = K = \begin{bmatrix} k_{11} & k_{12} & k_{13} \\ k_{21} & k_{22} & k_{23} \\ k_{31} & k_{32} & k_{33} \end{bmatrix} = concentration \\ matrix$ $K \in S^{3x3} = 3x3 \; real \; symmetric \; matrices.$. $K \in PD_3 = 3x3 \; positive \; definite \; matrices$. $\Xi = \{ a\begin{bmatrix} 1 & a & 1 \\ a & 1 & a \\ 1 & a & 1 \end{bmatrix}+ b \begin{bmatrix} 0 & 1 & 0 \\ 1 & 0 & 1 \\ 0 & 1 & 0 \end{bmatrix}:a, b \in \mathbb{R}\} \subset S^3$ $M= \{\varepsilon \sim N(0, \varepsilon) :\Sigma^{-1} \in \Xi \cap PD_3\}$ a linear concentration model. $M = Im \varphi (\Xi \cap PD_3) where$ $\varphi: S^3 - \rightarrow S^3$ $K \mapsto K^{-1}$ $\varphi (K) = {\sigma_{ij} = (-1)^{i+j}\tfrac{|K^{[3]}\backslash\{j\},[3]\backslash\{i\}|}{|K|}: i,j \in [3]}$ rational function in parameters a,b,... We need to extend our theorem to cover models with a rational parametrization. ### 50 * The may is then a rational map because each $\sigma_{ij}$ coordinate is specified by a rational function of the concentration parameters (i.e. a quotient of polynomials). * To haulle our statistical models we need to extend our solution of the implicitization problem to rational parametrizations. * $\textit{See example}$ To see the subtleties that arise lets consider an easier example of rational parametrization: $\varphi: \mathbb{R}^2 - \rightarrow \mathbb{R}^3$ $\varphi: (u,v) \rightarrow (\frac{u^2}{v}, \frac{v^2}{u}, u) = (x,y,z)$. * Note if $(x,y,z) \in Im (\varphi)$ then $x^2y - z^3 = 0$. * Try to apply polynomial implicitization by clearing denominators: $I = \langle vx - u^2, uy - v^2, z - u \rangle \subset \mathbb{Z} [u,v,x,y,z]$. *Exercise* $I_2 = \langle I \cap \mathbb{Z}[x,y,z] = x^2y - z^3 \rangle$. $\Rightarrow$ $V(I_2) = V (x^2y - z^3) \cup V (z)$ (see previous lemma!) $\Rightarrow$ $V(J_2)$ is not Zariski closure since $I'm (\varphi) \subset V (x^2y - z^2)$. * Need to remove $V(z)...$ ### 51 **General Set up:** $x_1 = \frac{f_1 (t_1, ..., t_m )}{{g_1 (t_1, ..., t_m )}}$ $\vdots$ $x_n = \frac{f_n (t_1, ..., t_m )}{{g_n (t_1, ..., t_m )}}$ Where $f_i, g_i, ..., f_n, gn \in \mathbb{K}[t_1, ..., t_m ]$ The map $\varphi :$ can not be defined on all of $\mathbb{K}^m$ since denominators cant be zero! **=>** have map $\varphi : \mathbb{K}^m / W \rightarrow \mathbb{K}^{n}$ $\varphi :( t_1, ..., t_m ) \mapsto ( \frac{f_1 (t_1, ..., t_m )}{g_1 (t_1, ..., t_m )},...,. \frac{f_n (t_1, ..., t_m )}{g_n (t_1, ..., t_m )})$ Where $W=V(g_1 g_2 ... g_n) \subset \mathbb{K}^{m}$ **Rational implicitization problem Find the smallest variety of $\mathbb{K}^{n}$ contiuing $\psi(\mathbb{K}^m /W)$** Adapt our maps: $\mathbb{K}^m / W \overset{\iota}{\longrightarrow} \mathbb{K}^{n+m}$ $\qquad \searrow \swarrow \pi_m$ $\qquad \qquad \mathbb{K}^n$ **Note** $i(\mathbb{K}^m / W ) \subset V(I)$, Where I is the ideal from clearing denominators** Add extra dimension to avoid vanishing denominators!: $G = G_1 ... G_n$ $J= \rangle g_1 ... g_n x_1- f_1..... G x_2- f_2, 1- gy \rangle \subset \mathbb{K}[y,t_t;t1...n ]$**(->) 1-gy => denominators do not vanish on V(J) !!** ### 52 Now have maps $\mathbb{K}^{n+m}$ $\qquad\searrow \nearrow \pi{m+1}$ $\mathbb{K}^{n} \rightarrow K^{n+m-1}$ Where $J:\mathbb{k}^{m}/{W} ➡️ Z^{m+} $\varphi (t_i ... t_m) ➡️ < ... \frac{H (t_k ... t_m)}{G ((t_i ... t_m)}$ Again we have $\phi = H + 1.0 * Then =3: Y = *^M +1 (0 => => G_ (ty =. Clain Prooh $2 (2M /N)$ $ 1+ 3m -G(t,....t_mX=F(T,.........1, 4 M7 => This implies 214,..... T_M22 =>-7, CAN 217 T0M2 Simce J=4 g(721 , From, 2 =.1 +102 and ( ( 27 (V) C V3 4(2M)= 22x+1(C V 3=M+(V3 IM (Y ) a WE IS Projection The Couse, 22N3 and Foun >T 1-7 the elinirnation Cousore theorens to Prove the fallowing ### 53 **Theorem (Rational Implicitization)** Let $\mathbb{K}$ be an infinite field, $\Psi : \mathbb{K}^m / W \to \mathbb{K}^n$ a rational parametrization above, and $J = \langle g, x-f, g-fn, gy-1\rangle c 214, where " = 44 , Then for " = , "41 (3), VC +1) is The Saallest alline Variety ertaining 444) Prool ( Exeries) Set on Analagus olf 9 soling the iuplicitization problew Example Canhinuiong Our Seall exaugle with neu Varicble We Sat = > ex > 24] s = Ja 24 "9,3 ]==22) * Vix"9-30 The Zamishti lasure of as expected! (Removed component Z) A subtlety haring = = we houe e Where 4-4", y) the 4M M the Soue 11 "eally compited V3" ( ### 54 ( (40) The lagis that will (4apually, a =2"M =U2 = 44. 2", that Cg(7) has Dimausia as 447 Then 470 S This aus=4""4-4,4 This will to apually herben for ss, sa dou't hory and sust ### **Exponential Familes** ##### 56 ## We now possess the Theory we Polyngmial Constraints and Camputahnal "ools Defining statistical * Cur model "algebrani statistical " "ealu" to Extract Models statistical - That admets Meaning soun "Models" Semialjebrate " " " 11 **+ to Suild up Cur Famaly of Cseuialgetrain statisticols * " W will start wr Th Exponential Foumilis*" Enponention Faniles are themost Commonly Studed / used statistical "Models"" EX A * * +Multivanate Morunal distributions* * +Multinourial Distobuhong+ + = Paison " - Cound Beta " Theit + + Tons, the the" the Foumilies have Enponational "2+22/6. Ex Erodent Desentt" (Gared "the ### (3):2=***K . &2 *771 1-20 is finite but it is & Finite beau > "11 24 (a) - "3+ the e a. - 6) + 1 V: 46 **=*** the - 1-20 is d 10 ### **&**7,4 53 is : * .T *the **44/ =L & e "the C 49,23) <2 the , --- ### **&**1 " &+ + " " 4& a "M +++ (94 " # **"'"'" " ```