Past exam of the mathematics course of the University of Cambridge / 2009 / iii / Paper 59 / 3

Set $d_0=0$ and assume $0<d_1<\cdots <d_m<n$. Choose the first member of the flag manifold as a $d_1$-plane, contributing $d_1(n-d_1)$ complex dimensions. After choosing $L_{i-1}$, choose $L_i/L_{i-1}$ as a $(d_i-d_{i-1})$-plane in $V/L_{i-1}$. Its Grassmannian has dimension $(d_i-d_{i-1})(n-d_i)$. Adding these fiber dimensions givesFor $V={\mathbb{C}}^4$ the results areHere $F_1={\mathbb{CP}}^3$ is projective twistor space, $F_2=\mathrm{Gr}(2,4)$ is complexified compactified Minkowski space, and $F_{12}$ is the twistor correspondence space of incident pairs consisting of a spacetime point and a twistor on its twistor line. Over $F_2$ its fiber is ${\mathbb{CP}}^1$; over $F_1$ its fiber is ${\mathbb{CP}}^2$, consistent with $4+1=3+2=5$. These are complex dimensions; the underlying real dimensions are $6,8,10$, while a real spacetime slice of $F_2$ has real dimension four.

Given two flags of the same dimensions, choose adapted ordered bases $u_1,\dots ,u_4$ and $v_1,\dots ,v_4$, so their first $d_i$ vectors span their respective $L_i$. The invertible map $A$ sending each $u_j$ to $v_j$ sends the first flag to the second. Let $\delta =\det A\ne 0$. Replace the last target basis vector by ${\delta }^{-1}{v}_4$; this does not alter any subspace in its flag. The corresponding map $g$ has determinant one and still sends every member of the first flag to the corresponding member of the second.

Thus the special linear group acts transitively on every fixed-type flag manifold:The adapted-basis construction is the needed extension of transitivity on nonzero vectors. For an arbitrary subgroup, vector transitivity alone would not justify transitivity on all flags; here the explicit determinant correction proves the stronger claim for the special linear group.

The special linear group acts on the twistor correspondence by $g\cdot (\ell ,W)=(g\ell ,gW)$. Since an invertible linear map preserves inclusion, this is again an incident flag. The two projections commute with that action:This is precisely equivariance: projecting a transformed flag gives the same result as transforming its projection. In particular $g$ sends the twistor line of a spacetime point to the line of the transformed point and sends the alpha-plane of a twistor to that of its transformed twistor. The incidence geometry is preserved, even when a transformation takes a point out of the chosen affine Minkowski chart.

Because $F_1$ is projective twistor space, the printed column represents the line $p=[e_1]$, not a distinguished nonzero vector. Its stabilizer subgroup consists of matrices whose first column is a nonzero multiple of $e_1$:The block form is necessary to preserve the line, and conversely every such determinant-one matrix preserves it. By transitivity, the orbit map $g\mapsto g[e_1]$ is onto $F_1$. Its two values agree precisely when $g^{-1}{g}^{\prime }\in H_p$, equivalently when $g{H}_p=g^{\prime }{H}_p$. This gives the homogeneous space identificationThe dimensions also agree: ${\dim }_{\mathbb{C}}SL(4)=15$, while the free $B$ and $b$ parameters give ${\dim }_{\mathbb{C}}{H}_p=9+3=12$; the quotient has dimension three.

This coset space is not a quotient group, because $H_p$ is not a normal subgroup. For example $h=I+E_{12}\in H_p$. Let $g$ have upper-left block $\left(\begin{array}{cc}0& -1\\ 1& 0\end{array}\right)$ and identity on the other two coordinates. Then $g\in SL(4,\mathbb{C})$ but $gh{g}^{-1}=I-E_{21}$ sends $e_1$ to $e_1-e_2$, so it is not in $H_p$. Thus multiplying coset representatives does not define a well-defined group operation.

If instead one required the literal vector $e_1$ to be fixed, one would have $a=1$ and $B\in SL(3,\mathbb{C})$; its quotient would be ${\mathbb{C}}^4\setminus \{0\}$, not $F_1$. The projective reading is required by the printed target quotient and the flag manifold definition.