Example of flip from linear algebra
Published:
$ \DeclareMathOperator{\Sym}{Sym} \DeclareMathOperator{\Spec}{Spec} \DeclareMathOperator{\Hom}{Hom} \def\sO{\mathscr{O}} \def\oto#1{\overset #1 \longrightarrow} \def\into{\hookrightarrow} \def\tensor{\otimes} \def\V{\mathbb{V}} \def\P{\mathbb{P}} \def\dual#1{#1 ^{\vee}} \def\iso{\cong} $
We give an example of flip using linear algebra to show that these surgery operations indeed are very common in algebraic geometry. Let $A, B$ be two vector spaces of dimensions $a$ and $b$ respectively, let’s say $b > a = 2$. We have natural map $A \times B \to A \otimes B$, sending $(a,b)$ to the simple otimes $a \otimes b$. The image of this map is precisely the set of simple tensors.
We note such situation is very common in algebraic geometry. Say $X$ is an algebraic variety, $L_1$ and $L_2$ are two line bundles on $X$, then we have the multplication map
\[H^0(X, L_1) \times H^0(X, L_2) \to H^0(X, L_1 \tensor L_2)\]the image is the set of divisors $D \in |L_2|$ that can be decomposed into a sum $D_1 + D_2$ with $D_i \in |L_i|$.
Consider $A \tensor B \iso \Hom(\dual{B}, A)$, where the simple tensors corresponds to linear maps $\dual{B} \to A$ of rank $1$. Consider the affine space $V(\Hom(\dual{B}, A)) = \Spec \Sym \Hom(\dual{B}, A)$, and consider the locus $\widetilde{X} \subseteq V(\Hom(\dual{B}, A))$ of linear maps of rank $\leq 1$. We want to study the geometry of $\widetilde{X}$.
First of all, we observe $\widetilde{X}$ has a single singular point at the origin. Because any linear map in $\Hom(\dual{B}, A)$ can be written as a $(2 \times b)$ matrix $(x_{ij})$ (after fixing a basis). Then it has rank $1$ if and only if it is nonzero, and all the $(2 \times 2)$ minors of $(x_{ij})$ vanish. This gives the equations cutting out $\widetilde{X} \subseteq V(\Hom(\dual{B}, A))$ from where we see $\widetilde{X}$ is singular at the origin. Furthermore the set of rank 1 matrices form a homogeneous space over $GL(2) \times GL(b)$. Hence if $\widetilde{X}$ were to be singular at a point other than the origin, it is singular everywhere. We conclude $\widetilde{X}$ is singular at the origin.
We will construct two natural resolutions of $\widetilde{X}$ giving a flip.
First of all, lets study $\widetilde{X}$ more closely. Recall from the first course in algebraic geometry, we learned in $\P(\dual{A} \otimes \dual{B})$, the space of simple tensors is precisely the Segre variety $X = \P(\dual{A}) \times \P(\dual{B}) \into \P(\dual{A} \otimes \dual{B})$. Hence $\widetilde{X}$ is precisely the cone over $X$ with vertex at the origin.
Secondly, observe giving a rank $1$ linear map $\varphi : \dual{B} \to A$ is the same as specifying a codimensional $1$ subspace $T \subseteq \dual{B}$ (being the kernel of $\varphi$), and a nonzero vector $a \in A$. This suggests us to consider the following vector bundle
\[V_1 := \V(\dual{A} \otimes \sO_{\P \dual{B}}(1)) \oto{\pi} \P \dual{B}\]note $\P \dual{B}$ parametrizes hyperplanes in $\P \dual{B}$, and over a hyperplane $T$, the fiber of $\pi$ is precisely the affine space $V(\dual{A})$, parametrizing vectors in $A$ (possibly the zero vector). Let $\iota : \P \dual{B} \into V_1$ denote the zero section.
By the description above, there is a natural map $V_1 \to \V(\dual{A} \otimes \dual{B})$ mapping onto $\widetilde{X}$. It is an isomorphism away from the origin in $\widetilde{X}$. Since $V_1$ is smooth, this gives a resolution of $\widetilde{X}$. The fiben over the singular point (the origin) is precisely the zero section $\iota(\P \dual{B}) \subseteq V_1$. Note since $\dim A = 2 > 1$, the zero section has codimension $2$, hence $V_1 \to \widetilde{X}$ is a small contraction.
Now we can repeat the same construction by switching $A$ and $B$: $A \otimes B \iso \Hom(\dual{B}, A) \iso \Hom(\dual{A}, B)$. We arrive at a second resolution $V_2 \to \widetilde{X}$ where
\[V_2 := \V(\dual{B} \otimes \sO_{\P \dual{A}}(1)) \oto{\pi} \P \dual{A}.\]We claim we have a flip:
\[\require{\AMScd} \begin{CD} @. V_2\\ @. @VVf_2V\\ V_1 @>>f_1> \widetilde{X} \end{CD}\]It suffices to show $-K_{V_1}$ is $f_1$-ample and $-K_{V_2}$ is $f_2$-ample. By Nakai-Moishezon it suffices to show $K_{V_1} . C < 0$ for any integral curve in $\iota(\P \dual{B})$ and $K_{V_2} . C > 0$ for any integral curve in $\iota(\P \dual{A})$.
We first compute $K_{V_1}$ and $K_{V_2}$. Recall if $E$ is a vector bundle on $X$, let $\V(E) := \Spec_X(\Sym E) \oto{\pi} X$ denote the total space of the vector bundle, then we have a exact sequence
\[\begin{CD} 0 @>>> \pi^* E^{\vee} @>>> T_{V} @>>>\pi^* T_X @>>> 0 \end{CD}\]By adjunction we conclude
\[K_V = \pi^* K_X + \det E.\]In our situation
\[\begin{align*} K_{V_1} = \pi^* K_{\P \dual{B}} + \sO_{\P \dual{B}}(2) = \pi^* \sO_{\P \dual{B}}(1-b),\\ K_{V_2} = \pi^* K_{\P \dual{A}} + \sO_{\P \dual{A}}(b) = \pi^* \sO_{\P \dual{A}}(b-2). \end{align*}\]Since $\iota(\P \dual{B})$ and $\iota(\P \dual{A})$ are projective spaces, there is only one extremal curve, namely a line. So it suffices to check the intersection with a line. Let $\ell \subseteq \P \dual{B}$ be a line, we have
\[\iota_* \ell . K_{V_1} = \ell . \iota^* K_{V_1} = \ell . \sO_{\P \dual{B}}(1-b) = 1 - b < 0\]and
\[\iota_* [\P \dual{A}] . K_{V_2} = \deg \iota^* K_{V_2} = \deg \sO_{\P \dual{A}}(b-2) = b - 2 > 0.\]We have a flip!