Vector bundles on rational homogeneous spaces

Given a smooth projective variety X and a vector bundle E on X, we denote ({mathcal {M}}) to be an unsplit family of rational curves on X. ({mathcal {M}}) is called unsplit if ({mathcal {M}}) is a proper ({mathbb {C}})-scheme. We say that E is uniform with respect to ({mathcal {M}}) if the restriction of E to the normalization of every curve in ({mathcal {M}}) splits as a direct sum of line bundles with the same splitting type. If X is a generalized Grassmannian (G/P_k), then we just call E uniform with respect to the special family of lines without mention the unsplit family ({mathcal {M}}).

Uniform vector bundles on generalized Grassmannians

Along this section we will work on uniform vector bundles on rational homogeneous spaces of Picard number one, i.e., generalized Grassmannians. Let G be a simple Lie group and (D={1,2,ldots ,n}) be the set of nodes of the Dynkin diagram ({mathcal {D}}) of G. Denote by (P_k) the parabolic subgroup of G corresponding to the node k. Consider the generalized Grassmannian ({mathcal {G}}=G/P_k) or, for brevity, ({mathcal {D}}/P_k). Denote by ({mathcal {M}}:=G/P_{N(k)}) the generalized flag manifold defined by the marked Dynkin diagram (({mathcal {D}},N(k))) and by ({mathcal {U}}:=G/P_{k,N(k)}=G/(P_{N(k)}cap P_k)) the universal family, which has a natural ({mathbb {P}}^{1})-bundle structure over ({mathcal {M}}), i.e., we have the natural diagram

Remarkably, ({mathcal {M}}) defined above is indeed an unsplit family of rational curves on ({mathcal {G}}). Given (xin {mathcal {G}}), ({mathcal {M}}_x=q(p^{-1}(x))), which we call the special family of lines of class (check{alpha _k}) through x, coincides with (H/P_{N(k)}) by Remark 2.7, where the set of nodes of the Dynkin diagram H is (D(H)=Dbackslash {k}).

Assume that k is an extremal node, that is, the subdiagram ({mathcal {D}}(H)) is connected. Remarkably, in the case ({mathcal {D}}=D_n), i.e., (G=SO(2n)), since (G/P_{n-1}cong G/P_n), we only need to think about the extremal node n. Similarly, since (E_6/P_1cong E_6/P_6), we just consider the extremal node 1 in (E_6). According to Theorem 2.6, ({mathcal {M}}_x) has the following possibilities:

• Projective spaces ({mathbb {P}}^{n}) or smooth quadrics (mathbf{Q} ^{n}),

• Grassmannians G(m, n),

• Spinor varieties ({mathcal {S}}_{n}),

• (E_6/P_6), (E_7/P_7), (C_3/P_3).

The possibilities are list in Table 2.

We observe that for (xin {mathcal {G}}), the set of morphisms from ({mathcal {M}}_x) to Grassmannians plays a critical role in determining whether a uniform vector bundle with respect to special family lines can split as a direct sum of line bundles. More precisely, let (varsigma) be a positive integer. As long as we show that every morphism ({mathcal {M}}_xrightarrow G(t,varsigma )) is constant for any integer (tge 1) and every (xin {mathcal {G}}), then uniform r-bundles with respect to special family lines on ({mathcal {G}}) split for (rle varsigma). We suggest that interested readers refer to Theorem 3.1 in paper [17] for details.

For the proof of Theorem 1.5, we will analyze the morphisms ({mathcal {M}}_xrightarrow G(t,varsigma )) one by one according to the probabilities of ({mathcal {M}}_x).

Case I. When ({mathcal {M}}_x) is a projective space ({mathbb {P}}^N) or a smooth quadric (mathbf{Q} ^N~(N=2m+1)), then their Chow rings have the form

where ({mathcal {H}}) is a hyperplane section. In particular, (text {dim} H^{2t}({mathcal {M}}_x,{mathbb {C}})=1) for every (1le tle N). By the proof of Lemma 3.4 in paper [17], every morphism ({mathcal {M}}_xrightarrow G(t,N)) is constant for any integer t.

When ({mathcal {M}}_x) is a smooth quadric (mathbf{Q} ^N~(N=2m)), since

where ({mathcal {H}}) is a hyperplane section and ({mathcal {U}}) is a subvariety of codimension m, then we get that the morphisms ({mathcal {M}}_xrightarrow G(t,N-1)) can only be constant similarly.

Case II. ({mathcal {M}}_x) is Grassmannian (G(d,n)~(dge 2)). We claim that every morphism (G(d,n)rightarrow G(t,n-d+1)) is constant for any integer (tge 1).

Lemma 3.1

There are nonconstant morphisms from (G(d,n)~(d ge 2)) to (G(t,n-d+1)) for any integer (tge 1).

Proof

Lemma 3.9 in paper [17] tells us that there are nonconstant morphisms from (G(2,n-d+2)) to (G(t,n-d+1)) for any integer (tge 1). Since G(d, n) is covered by (G(2,n-d+2)), there are nonconstant morphisms from G(d, n) to (G(t,n-d+1)) for any integer (tge 1). (square)

When ({mathcal {G}}=D_n/P_n) and ({mathcal {M}}_x=G(2,n)), by the above lemma, we obtain that every morphism (G(2,n)rightarrow G(t,n-1)) is constant for any integer (tge 1).

When ({mathcal {G}}=E_n/P_2~(n=6,7,8)) and ({mathcal {M}}_x=G(3,n)~(n=6,7,8)), by the above lemma, we obtain that every morphism (G(3,n)rightarrow G(t,n-2)~(n=6,7,8)) is constant for any integer (tge 1). Remarkably, in these cases, the value of (varsigma) can be appropriately enlarged. Since the dimension of (G(3,n)~(n=6,7,8)) is (3(n-3)), we can easily know

and

Since the Picard number of (G(3,n)~(n=6,7,8)) is one, the only morphisms

and

are all constant for any integer (tge 1).

Case III. ({mathcal {M}}_x) is spinor variety ({mathcal {S}}_n~(n=4,5,6)). The Chow ring of ({mathcal {S}}_n) is presented as a quotient of ({mathbb {Z}}[X_1,ldots ,X_n]) module the relations

for (1le sle n), where (X_j^{,}s) are the Schubert classes of codimension j, (X_0=1) and (X_j=0) for (j<0) or (j>n) (see Section 3.2 in [28]). In particular, (text {dim} H^{2t}({mathcal {M}}_x,{mathbb {C}})=1) for every (tle [frac{5}{2}]). Hence, the only morphisms ({mathcal {M}}_xrightarrow G(t,5)) are constant for any integer t. Remarkably, the value of (varsigma) can be appropriately enlarged. We will analyze ({mathcal {S}}_n~(n=4,5,6)) one by one and get the following conclusions.

1. (a)

   Claim: There is no nonconstant map from ({mathcal {S}}_4) to G(t, 7) for any integer (tge 1).

Proof: We can see the Chow ring of ({mathcal {S}}_4) is

1. (a.1)

   (1le tle 2). Since (text {dim} H^{2t}({mathcal {S}}_4,{mathbb {C}})=1), by the proof of Lemma 3.4 in paper [17], the only morphisms ({mathcal {S}}_4rightarrow G(t,7)) are constant.

2. (a.2)

   (t=3). Let (phi) be a morphism from ({mathcal {S}}_4) to G(3, 7). On ({mathcal {S}}_4), we have an exact sequence

   $$begin{aligned} 0rightarrow phi ^{*}H_3rightarrow {mathcal {O}}^{oplus 9}_{{mathcal {S}}_4}rightarrow phi ^{*}Q_4rightarrow 0. end{aligned}$$

   Then (c(phi ^{*}H_3)cdot c(phi ^{*}Q_4)=1). According to the Chow ring of ({mathcal {S}}_4), we can expand the equation into the following form:

   $$begin{aligned} (1+a_1X_1+a_2X_1^2+a_3X_1^3+&widetilde{a_3}X_3)cdot (1+b_1X_1+b_2X_1^2+\&b_3X_1^3+widetilde{b_3}X_3+b_4X_1^4+widetilde{b_4}X_1X_3)=1. end{aligned}$$

   By analyzing the coefficients on the left-hand side of the equation above, it’s easy to get (widetilde{a_3}=0). Therefore, this case can boil down to case (a.1).

3. (b)

   Claim: There is no nonconstant map from ({mathcal {S}}_5) to G(t, 9) for any integer (tge 1).

Proof: We can see the Chow ring of ({mathcal {S}}_5) is

1. (b.1)

   (1le tle 2). Since (text {dim} H^{2t}({mathcal {S}}_5,{mathbb {C}})=1), by the proof of Lemma 3.4 in paper [17], the only morphisms ({mathcal {S}}_5rightarrow G(t,9)) are constant.

2. (b.2)

   (t=3). Let (phi) be a morphism from ({mathcal {S}}_5) to G(3, 9). On ({mathcal {S}}_5), we have an exact sequence

   $$begin{aligned} 0rightarrow phi ^{*}H_3rightarrow {mathcal {O}}^{oplus 9}_{{mathcal {S}}_5}rightarrow phi ^{*}Q_6rightarrow 0. end{aligned}$$

   Then (c(phi ^{*}H_3)cdot c(phi ^{*}Q_6)=1). According to the Chow ring of ({mathcal {S}}_5), we can expand the equation into the following form:

   $$begin{aligned} (1+a_1X_1+&a_2X_1^2+a_3X_1^3+widetilde{a_3}X_3)cdot (1+b_1X_1+b_2X_1^2+b_3X_1^3+widetilde{b_3}X_3+b_4X_1^4+\&widetilde{b_4}X_1X_3+b_5X_1^5+widetilde{b_5}X_1^2X_3+b_5’X_5+b_6X_1^6+widetilde{b_6}X_1^3X_3+b_6’X_3^2)=1. end{aligned}$$

   Since (A^5({mathcal {S}}_5)) is freely generated over ({mathbb {Q}}) by the classes (X_1^5,X_1^2X_3) and (X_5), the above equation implies that the coefficient of (X_5) is 0, i.e., (b_5’=0). Since (A^6({mathcal {S}}_5)) and (A^7({mathcal {S}}_5)) are freely generated over ({mathbb {Q}}) by the classes ((X_1^6,X_1^3X_3,X_3^2)) and ((X_1^7,X_1^4X_3,X_1X_3^2)) respectively, the above equation implies that the coefficients of (X_3^2) and (X_1X_3^2) are all zero, i.e.,

   $$begin{aligned} b_6’+widetilde{a_3}widetilde{b_3}=0,~b_6’a_1+widetilde{a_3}widetilde{b_4}=0. end{aligned}$$

   It follows that

   $$begin{aligned} widetilde{a_3}widetilde{b_4}=widetilde{a_3}widetilde{b_3}a_1. end{aligned}$$

   (3.2)

   On the other hand, since (A^3({mathcal {S}}_5)) and (A^4({mathcal {S}}_5)) are freely generated over ({mathbb {Q}}) by the classes ((X_1^3,X_3)) and ((X_1^4,X_1X_3)), respectively, the above equation implies that the coefficients of (X_3) and (X_1X_3) are all zero, i.e.,

   $$begin{aligned} widetilde{a_3}+widetilde{b_3}=0,~widetilde{a_3}b_1+widetilde{b_3}a_1+widetilde{b_4}=0. end{aligned}$$

   (3.3)

   Combining the equations (3.2), (3.3) and (a_1=-b_1), we get (widetilde{a_3}=0). Therefore, this case can boil down to case (b.1).

3. (b.3)

   (t=4). By iterating the previous process, we get (c(phi ^{*}H_4)cdot c(phi ^{*}Q_5)=1), i.e.,

   $$begin{aligned} (1+a_1X_1+a_2X_1^2+a_3X_1^3+&widetilde{a_3}X_3+a_4X_1^4+widetilde{a_4}X_1X_3)cdot (1+b_1X_1+b_2X_1^2\&+b_3X_1^3+widetilde{b_3}X_3+b_4X_1^4+widetilde{b_4}X_1X_3+b_5X_1^5+widetilde{b_5}X_1^2X_3+b_5’X_5)=1. end{aligned}$$

   In a similar way, we can prove that (b_5’=widetilde{a_3}=widetilde{b_3}=0). Since (A^4({mathcal {S}}_5)) and (A^8({mathcal {S}}_5)) are freely generated over ({mathbb {Q}}) by the classes ((X_1^4,X_1X_3)) and ((X_1^8,X_1^5X_3,X_1^2X_3^2)), respectively, the above equation implies that the coefficients of (X_1X_3) and (X_1^2X_3^2) are all zero, i.e.,

   $$begin{aligned} widetilde{a_4}+widetilde{b_4}=0,~widetilde{a_4}widetilde{b_4}=0. end{aligned}$$

   Hence (widetilde{a_4}=0,widetilde{b_4}=0). Therefore, this case can also boil down to case (b.1).

4. (c)

   Claim: There is no nonconstant map from ({mathcal {S}}_6) to G(t, 11) for any integer (tge 1).

Proof: The Chow ring of ({mathcal {S}}_6) is presented as a quotient of ({mathbb {Z}}[X_1,X_3,X_5]) by the relations

1. (c.1)

   (1le tle 2). Since (text {dim} H^{2t}({mathcal {S}}_6,{mathbb {C}})=1), by the proof of Lemma 3.4 in paper [17], the only morphisms ({mathcal {S}}_6rightarrow G(t,11)) are constant.

2. (c.2)

   (t=3). Let (phi) be a morphism from ({mathcal {S}}_6) to G(3, 11). On ({mathcal {S}}_6), we have an exact sequence

   $$begin{aligned} 0rightarrow phi ^{*}H_3rightarrow {mathcal {O}}^{oplus 11}_{{mathcal {S}}_6}rightarrow phi ^{*}Q_8rightarrow 0. end{aligned}$$

   Then Then (c(phi ^{*}H_4)cdot c(phi ^{*}Q_5)=1). According to the Chow ring of ({mathcal {S}}_6), we can expand the equation into the following form:

   $$begin{aligned} (1+a_1X_1+&a_2X_1^2+a_3X_1^3+widetilde{a_3}X_3)cdot (1+b_1X_1+b_2X_1^2+b_3X_1^3+widetilde{b_3}X_3+b_4X_1^4+widetilde{b_4}X_1X_3\&+b_5X_1^5+widetilde{b_5}X_1^2X_3+b_5’X_5+b_6X_1^6+widetilde{b_6}X_1^3X_3+b_6’X_1X_5+b_6”X_3^2+b_7X_1^7\&+widetilde{b_7}X_1^4X_3+b_7’X_1^2X_5+b_7”X_1X_3^2+b_8X_1^8+widetilde{b_8}X_1^5X_3+b_8’X_1^3X_5+b_8”X_1^2X_3^2)=1. end{aligned}$$

   Since (A^3({mathcal {S}}_6)), (A^6({mathcal {S}}_6)) and (A^9({mathcal {S}}_6)) are freely generated over ({mathbb {Q}}) by the classes ((X_1^3,X_3)),

   ((X_1^6,X_1^3X_3,X_1X_5,X_3^2)) and ((X_1^9,X_1^6X_3,X_1^4X_5,X_1^3X_3^2,X_3^3)), respectively, the above equation implies that the coefficients of (X_3), (X_3^2) and (X_3^3) are all zero, i.e.,

   $$begin{aligned} widetilde{a_3}+widetilde{b_3}=0,~widetilde{a_3}widetilde{b_3}+b_6”=0,~widetilde{a_3}b_6”=0. end{aligned}$$

   Hence (widetilde{a_3}=widetilde{b_3}=b_6”=0). Therefore, this case can boil down to case (c.1).

3. (c.3)

   (t=4). By iterating the previous process, we get (c(phi ^{*}H_4)cdot c(phi ^{*}Q_7)=1), i.e.,

   $$begin{aligned} (1+a_1X_1+&a_2X_1^2+a_3X_1^3+widetilde{a_3}X_3+a_4X_1^4+widetilde{a_4}X_1X_3)cdot (1+b_1X_1+b_2X_1^2\&+b_3X_1^3+widetilde{b_3}X_3+b_4X_1^4+widetilde{b_4}X_1X_3+b_5X_1^5+widetilde{b_5}X_1^2X_3+b_5’X_5+b_6X_1^6\&+widetilde{b_6}X_1^3X_3+b_6’X_1X_5+b_6”X_3^2+b_7X_1^7+widetilde{b_7}X_1^4X_3+b_7’X_1^2X_5+b_7”X_1X_3^2)=1. end{aligned}$$

   In a similar way, we can prove (widetilde{a_3}=widetilde{b_3}=b_6”=0). Since (A^4({mathcal {S}}_6)), (A^8({mathcal {S}}_6)) and (A^{11}({mathcal {S}}_6)) are freely generated over ({mathbb {Q}}) by the classes

   $$begin{aligned} (X_1^4,X_1X_3), (X_1^8,X_1^5X_3,X_1^3X_5,X_1^2X_3^2)~ text {and}~ (X_1^{11},X_1^9X_3,X_1^6X_5,X_1^5X_3^2,X_1^2X_3^3) end{aligned}$$

   , respectively, the above equation implies that the coefficients of (X_1X_3), (X_1^2X_3^2) and (X_1^2X_3^3) are all zero, i.e.,

   $$begin{aligned} widetilde{a_4}+widetilde{b_4}=0,~widetilde{a_4}widetilde{b_4}+a_1b_7”=0,~widetilde{a_4}b_7”=0. end{aligned}$$

   It’s easy to check (widetilde{a_4}=0). Therefore, this case can boil down to case (c.1).

4. (c.4)

   (t=5). By iterating the previous process, we get (c(phi ^{*}H_5)cdot c(phi ^{*}Q_6)=1), i.e.,

   $$begin{aligned} (1+a_1X_1+&a_2X_1^2+a_3X_1^3+widetilde{a_3}X_3+a_4X_1^4+widetilde{a_4}X_1X_3+a_5X_1^5+widetilde{a_5}X_1^2X_3+a_5’X_5)\&cdot (1+b_1X_1+b_2X_1^2+b_3X_1^3+widetilde{b_3}X_3+b_4X_1^4+widetilde{b_4}X_1X_3+b_5X_1^5+\&widetilde{b_5}X_1^2X_3+b_5’X_5+b_6X_1^6+widetilde{b_6}X_1^3X_3+b_6’X_1X_5+b_6”X_3^2)=1. end{aligned}$$

   In a similar way, we can prove (widetilde{a_3}=widetilde{b_3}=b_6”=widetilde{a_4}=widetilde{b_4}=0). Since (A^5({mathcal {S}}_6)) and (A^{10}({mathcal {S}}_6)) are freely generated over ({mathbb {Q}}) by the classes ((X_1^5,X_1^2X_3,X_5)) and ((X_1^{10},X_1^7X_3,X_1^5X_5,X_1^4X_3^2,X_5^2)), respectively, the above equation implies that the coefficients of (X_5) and (X_5^2) are all zero, i.e.,

   $$begin{aligned} a_5’+b_5’=0,~a_5’b_5’=0. end{aligned}$$

   Hence (a_5’=b_5’=0). It follows that the coefficients of (X_1^2X_3) and (X_1^4X_3^2) are also zero, i.e.,

   $$begin{aligned} widetilde{a_5}+widetilde{b_5}=0,~widetilde{a_5}widetilde{b_5}=0. end{aligned}$$

   Hence (widetilde{a_5}=widetilde{b_5}=0). Therefore, this case can also boil down to case (c.1).

Case IV. ({mathcal {M}}_x=E_6/P_6). The Chow ring of (E_6/P_6) has the following form (see [7] Theorem 4). Let (y_1,y_4) be the Schubert classes on (E_6/P_6). Then

where

Lemma 3.2

There is no nonconstant map from (E_6/P_6) to G(t, 12) for any integer (tge 1).

Proof

1. (i).

   (1le tle 3). Since (text {dim} H^{2t}(E_6/P_6,{mathbb {C}})=1), by the proof of Lemma 3.4 in paper [17], the only morphisms (E_6/P_6rightarrow G(t,10)) are constant.

2. (ii).

   (4le tle 6). Let (phi) be a morphism from (E_6/P_6) to G(4, 12). On (E_6/P_6), we have an exact sequence

   $$begin{aligned} 0rightarrow phi ^{*}H_trightarrow {mathcal {O}}^{oplus 12}_{E_6/P_6}rightarrow phi ^{*}Q_{12-t}rightarrow 0. end{aligned}$$

   Then (c(phi ^{*}H_t)cdot c(phi ^{*}Q_{12-t})=1). Similar to Case III, we analyze the coefficients of the above equations and use Mathematica to conclude that these cases can boil down to case (i).

(square)

Case V. ({mathcal {M}}_x=E_7/P_7). The Chow ring of (E_7/P_7) has the following form (see [7] Theorem 6). Let (y_1,y_5) and (y_9) be the Schubert classes on (E_7/P_7). Then

where

Lemma 3.3

There is no nonconstant map from (E_7/P_7) to G(t, 14) for any integer (tge 1).

Proof

1. (i).

   (1le tle 4). Since (text {dim} H^{2t}(E_7/P_7,{mathbb {C}})=1), by the proof of Lemma 3.4 in paper [17], every morphism (E_7/P_7rightarrow G(t,14)) is constant.

2. (ii).

   (t=5). Let (phi) be a morphism from (E_7/P_7) to G(5, 14). On (E_7/P_7), we have an exact sequence

   $$begin{aligned} 0rightarrow phi ^{*}H_5rightarrow {mathcal {O}}^{oplus 14}_{E_7/P_7}rightarrow phi ^{*}Q_9rightarrow 0. end{aligned}$$

   Then (c(phi ^{*}H_5)cdot c(phi ^{*}Q_9)=1). According to the Chow ring of (E_7/P_7), we can expand the equation into the following form:

   $$begin{aligned} (1+a_1y_1+&a_2y_1^2+a_3y_1^3+a_4y_1^4+a_5y_1^5+widetilde{a_5}y_5)cdot (1+b_1y_1+b_2y_1^2+b_3y_1^3+b_4y_1^4+b_5y_1^5\&+widetilde{b_5}y_5+b_6y_1^6+widetilde{b_6}y_1y_5+b_7y_1^7+widetilde{b_7}y_1^2y_5+b_8y_1^8+widetilde{b_8}y_1^3y_5+b_9y_1^9+widetilde{b_9}y_1^4y_5+b_9’y_9)=1. end{aligned}$$

   Since (A^9(E_7/P_7)) is freely generated over ({mathbb {Q}}) by the classes (y_1^9), (y_1^4y_5) and (y_9), the above equation implies that the coefficient of (y_9) is 0, i.e., (b_9’=0). On this basis, notice that, (A^5(E_7/P_7)) and (A^{10}(E_7/P_7)) are freely generated over ({mathbb {Q}}) by the classes ((y_1^5,y_5)) and ((y_1^{10},y_1^5y_5,y_5^2)), respectively, the above equation implies that the coefficients of (y_5) and (y_5^2) are all zero, i.e.,

   $$begin{aligned} widetilde{a_5}+widetilde{b_5}=0,~widetilde{a_5}widetilde{b_5}=0. end{aligned}$$

   Hence (widetilde{a_5}=widetilde{b_5}=0). Therefore, this case can boil down to case (i).

3. (iii).

   (t=6). By iterating the previous process, we get (c(phi ^{*}H_6)cdot c(phi ^{*}Q_8)=1), i.e.,

   $$begin{aligned} (1+a_1y_1+&a_2y_1^2+a_3y_1^3+a_4y_1^4+a_5y_1^5+widetilde{a_5}y_5+a_6y_1^6+widetilde{a_6}y_1y_5)cdot (1+b_1y_1+b_2y_1^2+b_3y_1^3+\&b_4y_1^4+b_5y_1^5+widetilde{b_5}y_5+b_6y_1^6+widetilde{b_6}y_1y_5+b_7y_1^7+widetilde{b_7}y_1^2y_5+b_8y_1^8+widetilde{b_8}y_1^3y_5)=1. end{aligned}$$

   In a similar way, we can prove (widetilde{a_5}=widetilde{b_5}=0). Next, let’s consider the vanishing of (widetilde{a_6}). Since (A^6(E_7/P_7)) and (A^{12}(E_7/P_7)) are freely generated over ({mathbb {Q}}) by the classes ((y_1^6,y_1y_5)) and ((y_1^{12},y_1^2y_5^2,y_1^7y_5)), respectively, the above equation implies that the coefficients of (y_1y_5) and (y_1^2y_5^2) are all zero, i.e.,

   $$begin{aligned} widetilde{a_6}+widetilde{b_6}=widetilde{a_6}widetilde{b_6}=0. end{aligned}$$

   Hence, (widetilde{a_6}=widetilde{b_6}=0). Therefore, this case can boil down to case (i).

4. (iv).

   (t=7). By iterating the previous process, we get (c(phi ^{*}H_7)cdot c(phi ^{*}Q_7)=1), i.e.,

   $$begin{aligned} (1+a_1y_1+&a_2y_1^2+a_3y_1^3+a_4y_1^4+a_5y_1^5+widetilde{a_5}y_5+a_6y_1^6+widetilde{a_6}y_1y_5+a_7y_1^7+widetilde{a_7}y_1^2y_5)cdot (1+\&b_1y_1+b_2y_1^2+b_3y_1^3+b_4y_1^4+b_5y_1^5+widetilde{b_5}y_5+b_6y_1^6+widetilde{b_6}y_1y_5+b_7y_1^7+widetilde{b_7}y_1^2y_5)=1. end{aligned}$$

   In a similar way, we can prove (widetilde{a_5}=widetilde{b_5}=widetilde{a_6}=widetilde{b_6}=0). Next, let’s consider the vanishing of (widetilde{a_7}). Since (A^7(E_7/P_7)) and (A^{14}(E_7/P_7)) are freely generated over ({mathbb {Q}}) by the classes ((y_1^7,y_1^2y_5)) and ((y_1^{14},y_1^4y_5^2,y_1^9y_5)), respectively, the above equation implies that the coefficients of (y_1^2y_5) and (y_1^4y_5^2) are all zero, i.e.,

   $$begin{aligned} widetilde{a_7}+widetilde{b_7}=widetilde{a_7}widetilde{b_7}=0. end{aligned}$$

   Hence, (widetilde{a_7}=widetilde{b_7}=0). Therefore, this case can boil down to case (i).

(square)

Case VI. ({mathcal {M}}_x=C_3/P_3). The Chow ring of (C_3/P_3) is

where (y_1) and (y_3) are the Schubert classes on (C_3/P_3) (see Section 3.1 in [28]). Since (text {dim} H^{2t}(C_3/P_3,{mathbb {C}})=1) for (1le tle 2), every morphism (C_3/P_3rightarrow G(t,5)) is constant for any integer (tge 1).

Now we can summarize our proof as follows.

Proof of Theorem 1.5:

Theorem 1.5 follows from Lemmas 3.1, 3.2, 3.3 and the statements of Case I-VI above. (square)

Corollary 3.4

Consider the Dynkin diagrams in Sect. 2.

• If ({mathcal {G}}) is a generalized Grassmannian of type (A_n, B_n, C_n), or (D_n) with marked node k of degree 2, then

  $$begin{aligned} {mathcal {M}}_x=Z_1times Z_2, end{aligned}$$

  where (Z_1) is a generalized Grassmannian corresponding to the Dynkin sub-diagram consisting of nodes (1, 2,ldots , k-1) with marked node (k-1) and (Z_2) is a generalized Grassmannian corresponding to the Dynkin sub-diagram consisting of nodes (k+1, cdots , n) with marked node (k+1). Suppose (Z_1′) is a generalized Grassmannian corresponding to the Dynkin sub-diagram consisting of nodes (1, 2,ldots , k) with marked node k and (Z_2′) is a generalized Grassmannian corresponding to the Dynkin sub-diagram consisting of nodes (k, k+1,ldots , n) with marked node k. If

  $$begin{aligned} rle varsigma ({mathcal {G}}):=text {min}{varsigma (Z_1′), varsigma (Z_2′)}, end{aligned}$$

  then uniform r-bundles with respect to the special family of lines on ({mathcal {G}}) split as a direct sum of line bundles.

• If ({mathcal {G}}) is a generalized Grassmannian (E_7/P_5) or (E_8/P_6), then all uniform 2-bundles on ({mathcal {G}}) split as a direct sum of line bundles.

• If ({mathcal {G}}) is a generalized Grassmannian (E_8/P_5), then all uniform 2-bundles and 3-bundles on ({mathcal {G}}) split as a direct sum of line bundles.

Proof

If ({mathcal {G}}) is a generalized Grassmannian of type (A_n, B_n, C_n), or (D_n), in order to prove E splits, we just need to prove that every morphism ({mathcal {M}}_xrightarrow G(t,r)) is constant. Let (phi) be a morphism from ({mathcal {M}}_x) to G(t, r). Because every (Z_i~(1le ile 2)) can be regarded as the subvariety of ({mathcal {M}}_x) corresponding to the special family of lines of class (check{alpha _k}) through x on (Z_i’), we can consider the restriction of (phi) to (Z_i~(1le ile 2)). By assumption (rle text {min}{varsigma (Z_1′), varsigma (Z_2′)}), all the restriction maps are constant. Hence, (phi) is also constant. For the cases of (E_7) and (E_8), the proofs are same. (square)

Remark 3.5

• For the case ({mathcal {M}}_x={mathbb {P}}^1) or ({mathcal {M}}_x={mathbb {P}}^1times cdots), we can say nothing about the splitting result of other cases according to our theorem.

• In some cases, the value of (varsigma ({mathcal {G}})) cannot be big anymore, which means that there exist uniform but nonsplitting ((varsigma ({mathcal {G}})+1))-bundles. For instance, Grassmannian (A_{n-1}/P_d=G(d,n)~(dle n-d)) has uniform but nonsplitting d-bundle (H_d) (the universal bundle of G(d, n)); Spinor variety (D_{n+1}/P_{n+1}={mathcal {S}}_n) has uniform but nonsplitting ((n+1))-bundle (Q_{n+1}) (the universal quotient bundle of ({mathcal {S}}_n)).

• Compare to the main theorem (Theorem 3.1) in [17], we improve their results. In particular, we enlarge the splitting threshold for uniform bundles on Hermitian symmetric spaces (E_6/P_6) from (ge 5) to (ge 7) and (E_7/P_7) from (ge 7) to (ge 12) (see Table 1 in [17]).

It is known from Theorem 2.9 that any generalized Grassmannian ({mathcal {G}}) is covered by projective planes except for ({mathbb {P}}^1), (C_n/P_n) and (G_2/P_2). In order to test the splitting property of a vector bundle E on ({mathcal {G}}) covered by projective planes, one only needs to test the splitting property of E restricting to every projective plane ({mathbb {P}}^2subseteq {mathcal {G}}). Note that a vector bundle on ({mathbb {P}}^n) splits precisely when its restriction to some projective plane ({mathbb {P}}^2subseteq {mathbb {P}}^n) splits (see Theorem 2.3.2 in [23]). However, for a general generalized Grassmannian, we need the condition “restricting to every projective plane ({mathbb {P}}^2subseteq {mathcal {G}})” because the “positions” of ({mathbb {P}}^2)’s in a general generalized Grassmannian are different usually.

Corollary 3.6

Suppose ({mathcal {G}}) to be a generalized Grassmannian of the form (A_n/P_k~(nne 1)), (B_n/P_k~(kne n-1)), (C_n/P_k~(kne n)), (D_n/P_k), (E_n/P_k~(n=6,7,8)) or (F_4/P_k~(kne 2)). Let E be an r-bundle on ({mathcal {G}}). If E splits as a direct sum of line bundles when it restricts to every ({mathbb {P}}^2subseteq {mathcal {G}}), then E splits as a direct sum of line bundles on ({mathcal {G}}).

Proof

The condition implies that E is uniform with respect to the special family of lines. The reason for this is the fact every special line L is contained in two different ({mathbb {P}}^2) by Theorem 2.9.

We prove the corollary by induction on r. If we have the exact sequence of vector bundles

on ({mathcal {G}}), where the rank of M and N are both smaller than r, such that

for every (Zsimeq {mathbb {P}}^2), then by the induction hypothesis, M and N split. Since (H^1({mathcal {G}},N^{vee }otimes M)=0), the above exact sequence splits and hence also E.

Similar to the proof of Theorem 3.3 and Lemma 3.4 in [6], by using the notations in the diagram (3.1), on ({mathcal {U}}=G/(P_kcap P_{N(k)})), we can obtain an exact sequence

and the morphism

where ({mathbb {G}}(t-1,{mathbb {P}}(E^{vee }_x))) is the set of ((t-1))-dimensional projective subspaces of ({mathbb {P}}(E^{vee }_x)). If we prove that the morphism (varphi) is constant for every (xin {mathcal {G}}), then (widetilde{M}) and (widetilde{N}) are trivial on all p-fibers. So there exist two bundles M, N over ({mathcal {G}}) with (widetilde{M}={p}^{*}M,widetilde{N}={p}^{*}N). By projecting the bundle sequence

onto ({mathcal {G}}), we can get the desired exact sequence (3.7). Thus, to prove the existence of the above exact sequence, it suffices to show that every morphism

is constant for any (xin {mathcal {G}}).

Given a projective subspace Z of dimension 2 and a line (Lsubseteq Z), we take any point (xin L) and denote (Z’) to be the subspace of ({mathcal {M}}_x) corresponding to the tangent directions to Z at x. By the hypothesis, E|Z is a direct sum of line bundles, so every morphism

is constant. Since ({mathcal {G}}) is covered by linear projective subspaces of dimension 2 and ({mathcal {M}}_x) is chain-connected by ({mathbb {P}}^1) (see [20] Section 2.4), every morphism (varphi) is constant for any (xin {mathcal {G}}). (square)

Uniform vector bundles on rational homogeneous spaces

First, let’s recall the notations in Sect. 1.

Assume

where (G_i) is a simple Lie group with Dynkin diagram ({mathcal {D}}_i) whose set of nodes is (D_i) and (P_{I_i}) is a parabolic subgroup of (G_i) corresponding to (I_isubset D_i). We set (F(I_i):=G_i/P_{I_i}) by marking on the Dynkin diagram ({mathcal {D}}_i) of (G_i) the nodes corresponding to (I_i). Let (delta _i) be a node in ({mathcal {D}}_i) and (N(delta _i)) be the set of nodes in ({mathcal {D}}_i) that are connected to (delta _i).

If (delta _iin I_i),

is the i– th special family of lines of class (check{delta }_i) by Theorem 2.3, where (P_i^{delta _i^c}:=P_{(I_ibackslash delta _i)cup N(delta _i)}) and (widehat{G_i/P_{I_i}}) is (G_1/P_{I_1}times G_2/P_{I_2}times cdots times G_m/P_{I_m}) by deleting i-th term (G_i/P_{I_i}). For (i=1) and (delta in I), ({mathcal {M}}^{delta }) denotes the special family of lines of class (check{delta }).

For (xin X),

is the (delta _i)– th special part of variety of minimal rational tangents at x (sometimes we just write ({mathcal {M}}_x) if there is no confusion).

Fix (delta _iin I_i). ({mathcal {M}}_{i,x}^{delta _i^c}) is just the special family of lines of class (check{delta }_i) through x on the generalized Grassmannian ({mathcal {G}}^{delta _i}) whose Dynkin diagram ({mathcal {D}}^{delta _i}) is the maximal sub-diagram of (({mathcal {D}}_i,I_i)) with the only marked point (delta _i). Denote (nu (X, delta _i):=varsigma ({mathcal {G}}^{delta _i})) and

Proposition 3.7

Fix integer (i~(1le i le m)) and (delta _iin I_i). Let E be a uniform r-bundle on X of type ((a_1^{(delta _i)}, ldots ,a_r^{(delta _i)}),~a_1^{(delta _i)}le cdots le a_r^{(delta _i)}) with respect to ({mathcal {M}}_i^{delta _i^c}). If (rle nu (X,delta _i)) and these (a_j^{(delta _i)})’s are not all same, then E can be expressed as an extension of uniform bundles with respect to ({mathcal {M}}_i^{delta _i^c}). In particular, E is (delta _i)-unstable.

Proof

After twisting with an appropriate line bundle, we can assume that E has the splitting type

with respect to ({mathcal {M}}_i^{delta _i^c}).

Let’s consider the standard diagram

(3.8)

For (Lin {mathcal {M}}_i^{delta _i}), the (q_2)-fiber

is mapped under (q_1) to the line L identically in X and we have

For (xin X), the (q_1)-fiber ({q_1}^{-1}(x)) is mapped isomorphically under (q_2) to the subvariety

Because

for all (Lin {mathcal {M}}_i^{delta _i^c}). Thus the direct image ({q_2}_{*}{q_1}^{*}(E^{vee })) is a vector bundle of rank t over ({mathcal {M}}_i^{delta _i^c}). The canonical homomorphism of sheaves

makes (widetilde{N}^{vee }:={q_2}^{*}{q_2}_{*}{q_1}^{*}(E^{vee })) to be a subbundle of ({q_1}^{*}(E^{vee })). Because over each ({q_2})-fiber (widetilde{L}), the evaluation map

identifies (widetilde{N}^{vee }|_{widetilde{L}}) with ({mathcal {O}}_L^{oplus t}subseteq {mathcal {O}}_L^{oplus t}oplus bigoplus _{i=1}^{r-t} {mathcal {O}}_L(-a_{t+i}^{(delta _i)})=E^{vee }|_L.) Over ({mathcal {U}}_i^{delta _i^c}), we thus obtain an exact sequence

of vector bundles, whose restriction to ({q_2})-fibers (widetilde{L}) looks as follows:

Because (widetilde{N}^{vee }) is a subbundle of ({q_1}^{*}(E^{vee })) of rank t, for every point (xin X), it provides a morphism

where ({mathbb {G}}(t-1,{mathbb {P}}(E^{vee }_x))) is the set of ((t-1))-dimensional projective subspaces of ({mathbb {P}}(E^{vee }_x)). Since (rle nu (X,delta _i)), every morphism (varphi) is constant by the proof of Corollary 3.4 and the definition of (nu (X, delta _i)). It follows that (widetilde{M}) and (widetilde{N}) are trivial on all ({q_1})-fibers. So the canonical morphisms ({q_1}^{*}{q_1}_{*}widetilde{M}rightarrow widetilde{M}) and ({q_1}^{*}{q_1}_{*}widetilde{N}rightarrow widetilde{N}) are isomorphisms. Hence there are uniform bundles (M={q_1}_{*}widetilde{M}), (N={q_1}_{*}widetilde{N}) with respect to ({mathcal {M}}_i^{delta _i^c}) over X with

Then we obtain the exact sequence

by projecting the exact sequence (3.9) onto X. So (mu ^{(delta _i)}(N)>mu ^{(delta _i)}(E)) and thus E is (delta _i)-unstable (square)

Proposition 3.8

On X, if an r-bundle E is poly-uniform with respect to all the special families of lines such that the splitting type with respect to ({mathcal {M}}_i^{delta _i^c}) is ((a^{(delta _i)}, ldots , a^{(delta _i)})) for each i and (delta _i), then E splits as a direct sum of line bundles.

Proof

After twisting with an appropriate line bundle, we can assume that E is trivial on all the special families of lines on X. We are going to show that E is trivial.

Let’s first consider the case that X is a generalized flag manifold, which corresponds to a connected marked Dynkin diagram ({mathcal {D}}(X)) with l black nodes. We prove the lemma by induction on l. For (l=1), X is just a generalized Grassmannian. Then the result holds by Proposition 1.2 in [1]. Suppose the assertion is true for all generalized flag manifolds with connected marked Dynkin diagram and (l’) black nodes ((1le l'<l)). Let’s consider the natural projection

where (X’) is corresponding to the marked Dynkin diagram ({mathcal {D}}(X)) by changing the first black node (delta) to white. It’s not hard to see that every (pi)-fiber (pi ^{-1}(x)) is isomorphic to the generalized Grassmannian ({mathcal {G}}^{delta }) with the only marked point (delta). Since the restriction of E to every line in ({mathcal {M}}^{delta }) is trivial, so is to every line in (pi ^{-1}(x)). Thus E is trivial on all (pi)-fibers by Proposition 1.2 in [1]. It follows that (E’=pi _{*}E) is an algebraic vector bundle of rank r over (X’) and (Econg pi ^{*}E’).

Claim. (E’) is trivial on all the special families of lines on (X’).

In fact, let (gamma) be a black node in ({mathcal {D}}(X)) that is different from (delta) and L be a line in ({mathcal {M}}^{gamma }). Then (pi (L)) is also a line in (X’). When (gamma) runs through all black nodes except (delta) and L runs through all lines in ({mathcal {M}}^{gamma }) in X, (pi (L)) also runs through all lines in all the special families of lines of (X’). The projection (pi) induces an isomorphism

We identify L with (pi (L)). Since (E|_L) is trivial for every line L in ({mathcal {M}}^{gamma }) by assumption, (E’|_L) is trivial for every line L in all the special families of lines of (X’). By the induction hypothesis, (E’) is trivial. Thus (Econg q^{*}E’) is trivial.

Now let’s think about the general case that the marked Dynkin diagram of X is not connected. Assume X can be decomposed into a product

where (mge 2) and each (G_i~(1le ile m)) is a simple Lie group with a connected Dynkin diagram. We prove that E is trivial by induction on m. For (m=1), the result holds from the previous analysis. Considering the natural projection

one is easy to see that every f-fiber (f^{-1}(x)) is isomorphic to (G_1/P_{I_1}). By assumption, E is trivial on all the special families of lines on (f^{-1}(x)cong G_1/P_{I_1}). Thus E is trivial on all f-fibers by the previous analysis. It follows that (E’=f_{*}E) is an algebraic vector bundle of rank r over (X’) and (Econg f^{*}E’). Similarly, we can prove that (E’) is trivial and so is E. (square)

Theorem 3.9

On X, if r-bundle E is poly-uniform with respect to all the special families of lines and (rle nu (X)), then E is (delta _i)-unstable for some (delta _i) ((1le i le m)) or E splits as a direct sum of line bundles.

Proof

It is obviously from Proposition 3.7 and 3.8. (square)