Timeline for Coordinate free expression for the determinant of a $2 \times 2$ matrix
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| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Jan 27, 2019 at 19:15 | answer | added | Deane Yang | timeline score: 2 | |
| Jan 27, 2019 at 1:19 | comment | added | Deane Yang | This by the way is the Gauss equation that gives the Riemann curvature in terms of the second fundamental form of a hypersurface. Indeed, when the dimension is 2, the result is the determinant of the second fundamental form. | |
| Jan 26, 2019 at 18:54 | comment | added | Deane Yang | First, use the fact that $E\otimes E = \mathrm{Hom}(E^*, E)$. A map $h: E^* \rightarrow E$ induces a map $H: \bigwedge^2E^* \rightarrow \bigwedge^2E$, i.e., $H\in \bigwedge^2E\otimes \bigwedge^2E$. | |
| Jan 26, 2019 at 18:24 | comment | added | Sasha | 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. | |
| Jan 26, 2019 at 16:41 | comment | added | Jason Starr | 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? | |
| Jan 26, 2019 at 15:28 | comment | added | user61586 | No, I am in characteristic zero. I am thinking of a quadratic map $Sym^2(E)\rightarrow (\bigwedge^2 E)^{\otimes 2}$. But yes, frankly I am not even sure something like this exists. For instance, I tried starting with the map $E^{\otimes 4}\rightarrow (\bigwedge^2 E)^{\otimes 2}$ mapping $v_1\otimes v_2\otimes v_3\otimes v_4\mapsto (v_1\wedge v_3)\otimes (v_2\wedge v_4)+(v_2\wedge v_3)\otimes (v_1\wedge v_4)$ but it does not seem to induce a map $Sym^2(E)\rightarrow (\bigwedge^2 E)^{\otimes 2}$ with the property I want. | |
| Jan 26, 2019 at 15:22 | comment | added | Donu Arapura | 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 $GL_2$. | |
| Jan 26, 2019 at 15:20 | review | Close votes | |||
| Jan 28, 2019 at 9:38 | |||||
| Jan 26, 2019 at 15:17 | comment | added | user61586 | 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 $Sym^2(E)$. | |
| Jan 26, 2019 at 14:48 | comment | added | Ben McKay | 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? | |
| Jan 26, 2019 at 14:39 | history | asked | user61586 | CC BY-SA 4.0 |