Let $E\rightarrow X$ be a rank two vector bundle on a variety $X$. How can one write an abstract map $Sym^2(E)\rightarrow (\wedge^2 E)^{\otimes 2}$ that in local coordinates, when we fix a frame of $E$, gives the determinant of a $2\times 2$ symmetric matrix?
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1$\begingroup$ The vector bundle $\operatorname{Sym}^2E$ is spanned by local sections which are symmetric products $u \cdot v$. Map to $(u\wedge v)^{\otimes 2}$. Extend by linearity. Is that not the sort of answer you are looking for? $\endgroup$Ben McKay– Ben McKay2019-01-26 14:48:29 +00:00Commented Jan 26, 2019 at 14:48
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$\begingroup$ Thanks for the answer but not quite. In local coordinates write $u = ai+bj$ and $v = ci+dj$. Then $u\otimes v = ac (i\otimes i) + ad (i\otimes j) + bc (j\otimes i) + bd (j\otimes j)$, and the corresponding matrix would be $M = (m_{i,j})$ with $m_{1,1} = ac, m_{1,2} = ad, m_{2,1} = bc, m_{2,2} = bd$. Since the map I am looking for should give the determinant of the matrix it should be zero on tensors of type $u\otimes v$. Here i considered the non symmetric case but the same question make sense replacing $E\otimes E$ with $Sym^2(E)$. $\endgroup$user61586– user615862019-01-26 15:17:14 +00:00Commented Jan 26, 2019 at 15:17
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1$\begingroup$ I don't even think that there is a natural map unless you're in characteristic $2$. Otherwise, these correspond to distinct irreducible representations of $GL_2$. $\endgroup$Donu Arapura– Donu Arapura2019-01-26 15:22:32 +00:00Commented Jan 26, 2019 at 15:22
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1$\begingroup$ Why is there the tensor exponent "$\otimes 2$"? The usual determinant has target $\bigwedge^2 E$, not $(\bigwedge^2 E)^{\otimes 2}$. Are you asking about the square of the determinant? $\endgroup$Jason Starr– Jason Starr2019-01-26 16:41:22 +00:00Commented Jan 26, 2019 at 16:41
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4$\begingroup$ The determinant of a matrix with a fixed frame depends skew-symmetrically (not symmetrically) on the frame, so the source of your map should be $\wedge^2E$. Also, as @JasonStarr explained, the target is $\wedge^2E$. So, in the end, I think you are asking about a natural map $\wedge^2E \to \wedge^2E$. And the answer to this question is, of course, the identity. $\endgroup$Sasha– Sasha2019-01-26 18:24:31 +00:00Commented Jan 26, 2019 at 18:24
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1 Answer
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As mentioned in my comment above, the answer is given by the Gauss equations, which provide a formula for the Riemann curvature tensor in terms of the second fundamental form of a hypersurface in Euclidean space. Here is a description that is more explicit than the ones I gave in my comments above:
Let's view tensor products of $E$ to be multilinear functions of $E^*$. Then the map from $h \in S^2E$ to $R \in S^2\bigwedge^2 E$ is given by $$ R(v_1, v_2, v_3, v_4) = h(v_1,v_3)h(v_2,v_4) - h(v_1, v_4)h(v_2,v_3), $$ for any $v_1, v_2, v_3, v_4 \in E^*$. Note that, as a bonus, $R$ also satisfies the first Bianchi identities $$ R(v_1, v_2, v_3, v_4) + R(v_2, v_3, v_1, v_4) + R(v_3, v_1, v_2, v_4) = 0. $$
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$\begingroup$ Thanks a lot. I think this is what I am looking for. Say that $$h = \left(\begin{array}{cc} a & b \\ b & c \end{array} \right)$$ (as $2\times 2$ matrix), and set $v_i = (x_i,y_i)$. Then, if I got the computation right, we have $$R(v_1,v_2,v_3,v_4) = acx_1x_3y_2y_4-acx_1x_4y_2y_3-acx_2x_3y_1y_4+acx_2x_4y_1y_3-b^2x_1x_3y_2y_4+b^2x_1x_4y_2y_3+b^2x_2x_3y_1y_4-b^2x_2x_4y_1y_3$$ So the tensor $R$ is zero if and only the determinant of $h$ vanishes. Did I get your reasoning right? $\endgroup$user61586– user615862019-01-27 23:27:19 +00:00Commented Jan 27, 2019 at 23:27