Timeline for How few $k$-dimensional subspaces of $V$ are enough to have a complement to each $n-k$-dimensional subspace?
Current License: CC BY-SA 3.0
9 events
| when toggle format | what | by | license | comment | |
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| Oct 16, 2016 at 13:50 | comment | added | Julian Rosen | If $x^2 + ax + b$ is a quadratic polynomial without a root in $K$, we can take $D$ to be spanned by $(1,0,-a,1)$ and $(0,1,-b,0)$. The corresponding determinant is $\delta(e,f)=e^2+aef+bf^2$. So your argument does indeed apply to any field that isn't quadratically closed. | |
| Oct 16, 2016 at 13:40 | comment | added | YCor | @HeinrichD it's a little play with the action of $GL_4(K)$ on unordered triples of distinct 2-planes in $K^4$. One checks by hand that this action has very few orbits and in each case we explicitly see that it's not a complement repository. | |
| Oct 16, 2016 at 12:37 | comment | added | HeinrichD | How does one prove $d_K(4,2) \geq 4$? | |
| Oct 16, 2016 at 4:03 | history | edited | YCor | CC BY-SA 3.0 |
fixed typo added by mistake by another contributor + incorporated Sam's objection
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| Oct 16, 2016 at 2:52 | comment | added | darij grinberg | That's a beautiful argument! | |
| Oct 16, 2016 at 2:50 | history | edited | darij grinberg | CC BY-SA 3.0 |
various corrections
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| Oct 16, 2016 at 2:37 | comment | added | YCor | @SamHopkins Right; $d_K(4,2)\ge 4$ is an elementary exercise for an arbitrary field $K$ anyway. | |
| Oct 16, 2016 at 2:20 | comment | added | Sam Hopkins | This just shows $d_{\mathbb{R}}(4,2) \leq 4$, right? | |
| Oct 16, 2016 at 1:48 | history | answered | YCor | CC BY-SA 3.0 |