Prove
Lemma 2 ([Avoiding small collective images]). Let E,F be smooth Hermitian…Lemma 3 ([Continuous normalization]). Let E,F be Hermitian complex vector…Proposition
Fix the matrix size m and degree bound R . The bundles must satisfy two requirements. Their sum through level R must retain a Chern class after allowing the trivial directions coming from E_0 . At the same time, successive bundles must admit a fixed number of maps that expand every source subspace. We reduce each old line-bundle multiplicity by a factor of four at every step, so the total trivial part stays small. Each new projective-space factor is then sized to retain the Chern class contributed by every later occurrence of its line bundle. The initial rank leaves enough room in the reduced blocks to expand small subspaces; a large new block handles the others. The bundles and their base. For 1≤i≤R , define the fresh multiplicities b_i=(15)/(4)d_ i-1 . Let the multiplicity of the line bundle L_i in E_j be a_ 0,j , =d_0 4^ -j =64m4^ R-j ,; a_ i,j , =b_i4^ -(j-i) =960m4^ R+2i-j-2 (1≤i≤j≤R). All these numbers are positive integers. Put K_i=m∑_ j=i ^R a_ i,j =320m^2 4^ 2i-2 (4^ R-i+1 -1), Y=∏_ i=1 ^RCP^ K_i . Thus every K_i is a positive integer. Take L_0 to be the trivial line bundle on Y , and for i≥1 let L_i be the pullback of the hyperplane line bundle from the i th factor. Choose Hermitian metrics and define E_j=_ i=0 ^j L_i^ ⊕a_ i,j (0≤j≤R). In passing from E_j to E_ j+1 , each old multiplicity is divided by four and a fresh block of rank b_ j+1 =(15/4)d_j is added. Consequently the ranks satisfy rankE_ j+1 =14d_j+(15)/(4)d_j=4d_j, which proves the rank assertion by induction. Formula also shows that every multiplicity D to which division by four is applied is divisible by four and satisfies D≥256m . The retained Chern class. The choice of K_i records exactly the total multiplicity of L_i in the bundle to be tested. Set G=(_ j=0 ^R E_j)^ ⊕m , h_0=m∑_ j=0 ^R a_ 0,j , p=∑_ i=1 ^R K_i. The definitions of the multiplicities and K_i give G≅L_0^ ⊕h_0 ⊕ _ i=1 ^R L_i^ ⊕K_i , h_0=43md_0(1-4^ -(R+1) )<2md_0. Let z_i=c_1(L_i) . The integral cohomology ring is H^*(Y;ℤ) ≅ℤ[z_1,…,z_R]/ (z_1^ K_1+1 ,…,z_R^ K_R+1 ), z_i=2. In particular ∏_i z_i^ K_i is nonzero. The Whitney sum formula and the normalization for line bundles give c_p(G)=∏_ i=1 ^R z_i^ K_i ≠0; we use the standard Chern-class properties in . If G had a trivial subbundle of rank k>h_0 , its Hermitian complement would have rank h_0+p-k<p . The Whitney sum formula would identify c_p(G) with the p th Chern class of that complement, which vanishes above its rank. This contradiction, together with , proves. The argument applies to any continuous subbundle, without a condition on its position relative to the displayed line summands. The bases for the expanding maps. It remains to construct the maps between these bundles. For the step j→j+1 , with 0≤j<R , the source and target use only L_0,…,L_ j+1 and are pulled back from Y_ j+1 =∏_ i=1 ^ j+1 CP^ K_i . The dimensions K_i here are those in, including all later occurrences of L_i through level R . The complex dimension η_j of this smaller base satisfies η_j=∑_ i=1 ^ j+1 K_i , ≤43m∑_ i=1 ^ j+1 b_i; , =53md_0(4^ j+1 -1) <(20)/(3)md_j. Here the first inequality uses the geometric series with ratio 1/4 . We construct the maps for this step on Y_ j+1 , and then pull them back to Y . The estimate by a multiple of md_j is what will allow the incidence argument to handle all subspaces of dimension greater than m , even though the full base Y may be much larger. Sixteen maps for small subspaces. Decompose E_ j+1 as the sum of its replacement blocks and its fresh block. On a source block L_i^ ⊕D , the corresponding replacement block is L_i^ ⊕(D/4) . We seek 16 scalar matrices between the multiplicity spaces and tensor them with the identity of L_i . For 1≤s≤m , set r=D/4 and t=2s . Since D≥64m , t<r, r-2s≥D/8. The incidence expression of Lemma 2 ([Avoiding small collective images]), with q=16 and base a point, is strictly positive: (16s-2s)(D/4-2s)-s(D-s) =52sD-27s^2 ≥133ms>0. Applying Lemma 2 ([Avoiding small collective images]) simultaneously for s=1,…,m gives matrices whose images of every such s -plane span a space of dimension at least 2s . In fact, avoiding dimension at most 2s gives a slightly stronger conclusion. Make these choices for each block and assemble them into 16 block diagonal maps T_1,…,T_ 16 to the replacement part, with zero component in the fresh block. We verify that the same 16 maps expand every subspace of the whole source of dimension at most m . Work in one fiber and write the ordered source and replacement blocks as V_k and F_k . Let V_ ≤k and F_ ≤k denote their initial sums. Given a subspace S of dimension at most m , set S_k=S∩V_ ≤k , W=∑_ i=1 ^ 16 T_iS, W_k=W∩F_ ≤k . Use the zero spaces before the first flag terms. Projection onto the k th block identifies S_k/S_ k-1 and W_k/W_ k-1 with subspaces of V_k and F_k , respectively. Block diagonality gives T_i(S_k)⊆W_k and hence ∑_ i=1 ^ 16 T_ i,k (S_k/S_ k-1 ) ⊆W_k/W_ k-1 , where T_ i,k is the map on the k th block. Each source quotient has dimension at most m , so the block estimates yield W =∑_k(W_k/W_ k-1 ) ≥2∑_k(S_k/S_ k-1 ) =2 S. This proves the required estimate even when S has correlated components in several blocks. Eight maps for large subspaces. Let d=d_j and let the target now be the fresh block of rank r=(15/4)d . For every integer m<s≤d , take q=8 and t=2s . The rank endpoint is valid because 2s≤2d<r . The incidence expression satisfies (8s-2s)(15d/4-2s)-s(d-s) , =(43)/(2)sd-11s^2; , ≥(21)/(2)sd >(20)/(3)md >η_j. The first inequality follows from s≤d . Lemma 2 ([Avoiding small collective images]) on Y_ j+1 , applied to the finite list s=m+1,…,d , therefore supplies eight smooth maps to the fresh block whose image spans have dimension at least 2s for all these subspaces. Regard these as maps to E_ j+1 with zero replacement components. Together, the two lists give 24 maps T_i:E_j→E_ j+1 satisfying ∑_ i=1 ^ 24 T_iS≥2 S for every fiber subspace S . The estimate is supplied by the first list when S≤m and by the second when S>m ; adding maps cannot decrease an image span. Pull back to Y and apply Lemma 3 ([Continuous normalization]). This gives continuous maps X_i^ (j) satisfying for every 0≤j<R . The scaling may mix the displayed line summands, but each resulting map still has source E_j and target E_ j+1 . Continuation beyond the test. For j≥R , define E_ j+1 =E_j⊕E_j , and let X_1^ (j) =2^ -1/2 ι_1, X_2^ (j) =2^ -1/2 ι_2, X_i^ (j) =0 (3≤i≤24), where ι_1,ι_2 are the two inclusions. Their column sum is I_ E_j and their row sum is 12I_ E_ j+1 , so holds for these steps as well. This continuation leaves the retained bundle in unchanged and completes the construction.