Theorem 2.1 Set $k := \lfloor n/2 \rfloor$. The collection of $2k + 1$ line bundles

(2.2)\begin{equation} \mathcal{A} := \langle \mathcal{O}(0,0), \mathcal{O}(1,0), \mathcal{O}(0,1), \mathcal{O}(2,0), \mathcal{O}(0,2),\ldots, \mathcal{O}(k,0), \mathcal{O}(0,k) \rangle \end{equation}

in ${{\mathbf {D}^{\mathrm {b}}}}(X)$ is exceptional and extends to an $\operatorname {Aut}(X)$-invariant semiorthogonal decomposition

(2.3)\begin{equation} {{\mathbf{D}^{\mathrm{b}}}}(X) = \langle \mathcal{R}, \mathcal{A}, \mathcal{A} \otimes \mathcal{O}(1,1),\ldots, \mathcal{A} \otimes \mathcal{O}(n-1,n-1) \rangle \end{equation}

where $\mathcal {R}$ is the residual category.

1. (1) If $n = 2k$ the residual category is zero.

2. (2) If $n = 2k + 1$ the residual category is generated by the $\operatorname {Aut}(X)$-invariant exceptional collection

   (2.4)\begin{align} \mathcal{R} = \langle \mathcal{O}({-}1,k), &\mathcal{O}(k,-1); \mathcal{E}({-}1,k-1), \mathcal{E}^{{\vee}}(k-1,-1);\nonumber\\ &\ldots; \Lambda^{k-1}\mathcal{E}({-}1,1), \Lambda^{k-1}\mathcal{E}^{{\vee}}(1,-1); \Lambda^{k}\mathcal{E}({-}1,0) \cong \Lambda^{k}\mathcal{E}^{{\vee}}(0,-1) \rangle \end{align}
   of length $n = 2k + 1$ and is equivalent to the derived category of the Dynkin quiver $\mathrm {A}_n$.

Proof. Consider the standard exact sequence

\[ 0 \to \mathcal{O}_{{{\mathbb{P}}}(V) \times {{\mathbb{P}}}(V^{{\vee}})}(a-1,b-1) \to \mathcal{O}_{{{\mathbb{P}}}(V) \times {{\mathbb{P}}}(V^{{\vee}})}(a,b) \to \mathcal{O}(a,b) \to 0. \]

If any of the conditions listed in the lemma is satisfied, then

\[ H^{{\bullet}}({{\mathbb{P}}}(V) \times {{\mathbb{P}}}(V^{{\vee}}), \mathcal{O}_{{{\mathbb{P}}}(V) \times {{\mathbb{P}}}(V^{{\vee}})}(a,b)) \quad\text{and}\quad H^{{\bullet}}({{\mathbb{P}}}(V) \times {{\mathbb{P}}}(V^{{\vee}}), \mathcal{O}_{{{\mathbb{P}}}(V) \times {{\mathbb{P}}}(V^{{\vee}})}(a-1,b-1)) \]

vanish, hence $H^{\bullet }(X,\mathcal {O}(a,b))$ vanishes as well.

Proof. To prove the corollary we need to show that

\[ \operatorname{Ext}^{{\bullet}}(\mathcal{O}(j+t,t),\mathcal{O}(i,0)) = H^{{\bullet}}(X,\mathcal{O}(i-j-t,-t)) \]

and

\[ \operatorname{Ext}^{{\bullet}}(\mathcal{O}(t,j+t),\mathcal{O}(i,0)) = H^{{\bullet}}(X,\mathcal{O}(i-t,-j-t)) \]

both vanish when $0 \le i,j \le k$ and $1 \le t \le n-1$.

The first vanishing is clear since $1-n \le -t \le -1$ and Lemma 2.2 applies. The second vanishing also follows if $-j-t \ge 1-n$. So, assume $-j-t \le -n$. Then $t \ge n - j \ge n - k$, hence $i - t \le k - (n - k) = 2k - n \le 0$. Since also $i - t \ge -t \ge 1 - n$, the vanishing also follows from Lemma 2.2, unless $i - t = 0$. But then we must have $i = j = t = k$ and $n = 2k$, so that the corresponding line bundle is $\mathcal {O}(0,-n)$, and its cohomology vanishes, again by Lemma 2.2.

Proof. The ${{\mathbb {P}}}^{2k-1}$-fibration $p_1$ gives rise to the semiorthogonal decomposition

\[ {{\mathbf{D}^{\mathrm{b}}}}(X) = \big\langle p_1^{*}({{\mathbf{D}^{\mathrm{b}}}}({{\mathbb{P}}}(V))), p_1^{*}({{\mathbf{D}^{\mathrm{b}}}}({{\mathbb{P}}}(V))) \otimes \mathcal{O}(0,1),\ldots, p_1^{*}({{\mathbf{D}^{\mathrm{b}}}}({{\mathbb{P}}}(V))) \otimes \mathcal{O}(0,2k-1) \big\rangle. \]

Choosing the exceptional collection

\[ {{\mathbf{D}^{\mathrm{b}}}}({{\mathbb{P}}}(V)) = \langle \mathcal{O}(i-k), \mathcal{O}(i-k+1),\ldots, \mathcal{O}(i+k) \rangle \]

in the $i$th component, we obtain a full exceptional collection in ${{\mathbf {D}^{\mathrm {b}}}}(X)$ that takes the form

(2.5)\begin{equation} \begin{aligned} {{\mathbf{D}^{\mathrm{b}}}}(X) = \langle & \mathcal{O}({-}k,0), \mathcal{O}(1-k,0),\ldots, \mathcal{O}(k,0), \nonumber\\ & \quad\mathcal{O}(1-k,1), \mathcal{O}(2-k,1),\ldots, \mathcal{O}(k+1,1), \\ & \quad\quad\hspace{8em} \ldots\nonumber\\ & \quad\quad\quad\mathcal{O}(k-1,2k-1), \mathcal{O}(k,2k-1),\ldots, \mathcal{O}(3k-1,2k-1)\rangle.\end{aligned}\end{equation}

The collection is shown in Figure 1, the objects are represented by black dots.

Figure 1. Mutation of (2.5) to a rectangular Lefschetz collection for $k = 3$.

Now we perform a mutation: we consider the subcollection formed by the first $k$ terms of the first line, the first $k-1$ terms of the second line, and so on, up to the first term of the $k$th line of (2.5):

(2.6)\begin{equation} \begin{aligned} \{ \mathcal{O}({-}k,0), \mathcal{O}(1-k,0),\ldots, & \mathcal{O}({-}1,0); \\ \mathcal{O}(1-k,1),\ldots, & \mathcal{O}({-}1,1); \\ \ddots \hphantom{,} & \qquad\vdots \\ & \mathcal{O}({-}1,k-1) \} \end{aligned} \end{equation}

(this subcollection is depicted by the triangle on the left in Figure 1) and mutate it to the far right of the exceptional collection. Using Lemma 2.2, it is easy to see that the objects in (2.6) are right-orthogonal to the objects in the rest of (2.5). Hence, their mutation to the far right is realized by the anticanonical twist (see, for example, [Reference Ingalls and KuznetsovIK15, Lemma 2.13]). As $\omega _X^{-1} \cong \mathcal {O}(2k,2k)$, this replaces the black dots in the triangle by the white dots in the dashed triangle in Figure 1.

All that remains is to note that the resulting full exceptional collection is precisely the collection

\[ \langle \mathcal{A}, \mathcal{A} \otimes \mathcal{O}(1,1),\ldots, \mathcal{A} \otimes \mathcal{O}(2k-1,2k-1) \rangle. \]

Indeed, the blocks formed by the twists of the subcategory $\mathcal {A}$ correspond to the L-shapes in figure, and their semiorthogonality was proved in Corollary 2.3.

Proof. The ${{\mathbb {P}}}^{2k}$-fibration $p_1$ gives rise to the semiorthogonal decomposition

\[ {{\mathbf{D}^{\mathrm{b}}}}(X) = \big\langle p_1^{*}({{\mathbf{D}^{\mathrm{b}}}}({{\mathbb{P}}}(V))), p_1^{*}({{\mathbf{D}^{\mathrm{b}}}}({{\mathbb{P}}}(V))) \otimes \mathcal{O}(0,1),\ldots, p_1^{*}({{\mathbf{D}^{\mathrm{b}}}}({{\mathbb{P}}}(V))) \otimes \mathcal{O}(0,2k) \big\rangle. \]

This time for the first $k$ components (i.e. for $0 \le i \le k-1$) we choose the collection

\[ {{\mathbf{D}^{\mathrm{b}}}}({{\mathbb{P}}}(V)) = \langle \mathcal{O}(i-k), \mathcal{O}(i-k+1),\ldots, \mathcal{O}(i+k+1) \rangle, \]

and for the last $k + 1$ components (i.e. for $k \le i \le 2k$) we choose the collection

\[ {{\mathbf{D}^{\mathrm{b}}}}({{\mathbb{P}}}(V)) = \langle \mathcal{O}(i-k-1), \mathcal{O}(i-k),\ldots, \mathcal{O}(i+k) \rangle. \]

As a result, we obtain (2.7).

Proof. It follows from Lemma 2.2 that the object $\mathcal {O}(-1,k)$ (corresponding to the circled grey cross (red online) in Figure 2) is already in the residual category, so it is enough to mutate the other $2k$ objects.

Consider the exact sequences

(2.12)\begin{align} 0 \to \Omega^{k+i}(k+i) \boxtimes_X \mathcal{O}(i-1) \to &\Lambda^{k+i}V^{{\vee}} \otimes \mathcal{O}(0,i-1) \to \nonumber\\ & \cdots\to V^{{\vee}} \otimes \mathcal{O}(k+i-1,i-1) \to \mathcal{O}(k+i,i-1) \to 0 \end{align}

obtained by an appropriate twist of the pullback to $X$ of the truncated Koszul complex on ${{\mathbb {P}}}(V)$, and analogous exact sequences

(2.13)\begin{align} 0 \to \mathcal{O}(i-1) \boxtimes_X \Omega^{k+i}(k+i) \to &\Lambda^{k+i}V \otimes \mathcal{O}(i-1,0) \to \nonumber\\ & \cdots\to V \otimes \mathcal{O}(i-1,k+i-1) \to \mathcal{O}(i-1,k+i) \to 0. \end{align}

If we show that the objects $\Omega ^{k+i}(k+i) \boxtimes _X \mathcal {O}(i-1)$ and $\mathcal {O}(i-1) \boxtimes _X \Omega ^{k+i}(k+i)$ belong to $\mathcal {R}$, it will follow that these exact sequences express mutations of $\mathcal {O}(k+i,i-1)$ and $\mathcal {O}(i-1,k+i)$ through the rectangular part of (2.3), and that together with the object $\mathcal {O}(-1,k)$ these objects generate the residual category $\mathcal {R}$.

So, we need to check that $\operatorname {Ext}^{\bullet }(\mathcal {O}(a,b),-) = H^{\bullet }(X, - \otimes \mathcal {O}(-a,-b))$ vanishes on these objects when $\mathcal {O}(a,b)$ run through the set of objects generating the rectangular part of (2.3), that is, for

(2.14)\begin{equation} 0 \le a \le b \le a + k \le 3k \quad \text{and} \quad 0 \le b \le a \le b + k \le 3k. \end{equation}

We have exact sequences on ${{\mathbb {P}}}(V) \times {{\mathbb {P}}}(V^{\vee })$,

\[ 0 \to \Omega^{k+i}(k+i-1) \boxtimes \mathcal{O}(i-2) \to \Omega^{k+i}(k+i) \boxtimes \mathcal{O}(i-1) \to \Omega^{k+i}(k+i) \boxtimes_X \mathcal{O}(i-1) \to 0. \]

To compute $H^{\bullet }(X,\Omega ^{k+i}(k+i-a) \boxtimes _X \mathcal {O}(i-1-b))$ we thus need to compute two tensor products:

(2.15)\begin{equation} \begin{aligned} H^{{\bullet}}({{\mathbb{P}}}(V), \Omega^{k+i}(k+i-a)) & \otimes H^{{\bullet}}({{\mathbb{P}}}(V^{{\vee}}), \mathcal{O}(i-1-b)), \\ H^{{\bullet}}({{\mathbb{P}}}(V), \Omega^{k+i}(k+i-1-a)) & \otimes H^{{\bullet}}({{\mathbb{P}}}(V^{{\vee}}), \mathcal{O}(i-2-b)). \end{aligned} \end{equation}

By Bott's formula [Reference BottBot57, Proposition 14.4] the first factors in these products are zero, except for

(2.16)\begin{equation} \begin{aligned} & a \le -1, && \text{or} && a = k+i, && \text{or} && a \ge 2k + 2, \\ & a \le -2, && \text{or} && a = k+i-1, && \text{or} && a \ge 2k + 1 \end{aligned} \end{equation}

(the conditions in the first (respectively, second) line of (2.16) corresponds to non-vanishing of the first factors in the first (respectively, second) line in (2.15)). Similarly, the second factors vanish except for

(2.17)\begin{equation} \begin{aligned} & b \le i - 1, && \text{or} && b \ge 2k + i + 1, \\ & b \le i - 2, && \text{or} && b \ge 2k + i \end{aligned}\end{equation}

(with the same convention about the role of the lines). Clearly, (2.14) is not compatible with the conditions $a \le -1$ or $a \le -2$ in (2.16). Furthermore, (2.14) implies that $|a - b| \le k$, which contradicts all conditions in (2.16) and (2.17), except for those in the last columns. However, in the latter case both $a$ and $b$ are strictly bigger than $2k$ (recall that $1 \le i \le k$), which also contradicts (2.14).

Thus, we conclude that for all $(a,b)$ satisfying (2.14) either of the factors in the tensor products (2.15) vanishes, hence the objects $\Omega ^{k+i}(k+i) \boxtimes _X \mathcal {O}(i-1)$ belong to the residual category $\mathcal {R}$. By symmetry it also follows that the objects $\mathcal {O}(i-1) \boxtimes _X \Omega ^{k+i}(k+i)$ belong to $\mathcal {R}$. Thus, all the objects on the right-hand side of (2.10) are in $\mathcal {R}$. Moreover, as we already pointed out, it also follows that (2.12) and (2.13) are mutation sequences. Therefore, together with the line bundle $\mathcal {O}(-1,k)$ the objects on the right-hand side of (2.10) generate $\mathcal {R}$.

Similarly, from Lemma 2.2 and the mutation sequences, orthogonality between the rows of (2.10) follows, so all that remains is to compute $\operatorname {Ext}$-spaces between objects in each row as well as from objects in the first row to $\mathcal {O}(-1,k)$, and from $\mathcal {O}(-1,k)$ to objects in the last row.

For this we use (2.12) and (2.13) as resolutions for the source objects. Since almost all terms in these sequences are in the rectangular part of (2.3), it follows that it is enough to compute $\operatorname {Ext}$-spaces from the rightmost objects only (i.e. for the grey crosses (red online) from Figure 2). Thus, we need to describe the tensor products in (2.15) for $(a,b)$ satisfying

\[ 0 \le a \le k - 1,\quad b = a + k + 1\quad\text{or}\quad 0 \le b \le k - 1,\quad a = b + k + 1. \]

Clearly, the assumption $0 \le a \le k - 1$ contradicts all conditions in (2.16). On the other hand, if $0 \le b \le k - 1$ and $a = b + k + 1$, then $a \le 2k$, so the only compatible conditions are

\[ (a,b) = (k + i, i - 1) \quad\text{or}\quad (a,b) = (k + i - 1, i - 2). \]

Furthermore, in the first of these cases the first tensor product in (2.15) is one-dimensional (and lives in cohomological degree $k+i$), and in the second of these cases the second tensor product is one-dimensional (and lives in the same cohomological degree). This computation proves that the only $\operatorname {Ext}$-spaces between the objects in the first and the last rows of (2.10) are those listed in the first and last lines of (2.11).

Next, we compute $\operatorname {Ext}$-spaces from objects in the first row of (2.10) to $\mathcal {O}(-1,k)$. As before, we use (2.12) as resolutions. The only object that appears in (2.12) which is not semiorthogonal to $\mathcal {O}(-1,k)$ is $\mathcal {O}(2k,k-1)$ (see Lemma 2.2 and (2.8)), from which we have a one-dimensional $\operatorname {Ext}$-space in degree $2k$. Therefore, the only $\operatorname {Ext}$-space from objects of the first row to $\mathcal {O}(-1,k)$ is given by the second line of (2.11).

Finally, we compute $\operatorname {Ext}$-spaces from $\mathcal {O}(-1,k)$ to objects in the last row of (2.10). First, we consider the Koszul exact sequence

\[ 0 \to \mathcal{O}({-}1,k) \to \Lambda^{2k+1}V^{{\vee}} \otimes \mathcal{O}(0,k) \to \cdots \to V^{{\vee}} \otimes \mathcal{O}(2k,k) \to \mathcal{O}(2k+1,k) \to 0. \]

All its terms (except for the leftmost and rightmost) are in the rectangular part of (2.3), hence

\[ \operatorname{Ext}^{p}(\mathcal{O}({-}1,k), \mathcal{O}(i-1) \boxtimes_X \Omega^{k+i}(k+i)) \cong \operatorname{Ext}^{p + 2k + 1}(\mathcal{O}(2k+1,k), \mathcal{O}(i-1) \boxtimes_X \Omega^{k+i}(k+i)). \]

Now we use (2.13) as a resolution for $\mathcal {O}(i-1) \boxtimes _X \Omega ^{k+i}(k+i)$. Clearly, the only non-trivial $\operatorname {Ext}$-space from $\mathcal {O}(2k+1,k)$ to its terms is

\[ \operatorname{Ext}^{2k}(\mathcal{O}(2k+1,k),\mathcal{O}(0,k+1)) \cong H^{2k}(X,\mathcal{O}({-}2k-1,1)) \cong \Bbbk \]

(which holds for $i = k + 1$). Therefore, the only $\operatorname {Ext}$-space from $\mathcal {O}(-1,k)$ to the last row of (2.3) is given by the third line of (2.11).

It is clear from the above description of $\operatorname {Ext}$-spaces that the category $\mathcal {R}$ is equivalent to the derived category of the Dynkin quiver $\mathrm {A}_{2k+1}$ (with the objects in (2.10) corresponding to simple representations of the quiver, up to shift).

Proof. The definition (2.1) of the object $\mathcal {E}$ can be rewritten as the exact sequence

(2.18)\begin{equation} 0 \to \mathcal{E} \to \mathrm{T}({-}1) \boxtimes_X \mathcal{O} \to \mathcal{O}(0,1) \to 0. \end{equation}

It follows that $\operatorname {rk}(\mathcal {E}) = 2k$ and $\det (\mathcal {E}) \cong \mathcal {O}(1,-1)$, hence

(2.19)\begin{equation} \Lambda^{i}\mathcal{E}^{{\vee}} \cong \Lambda^{2k-i}\mathcal{E} \otimes \mathcal{O}({-}1,1). \end{equation}

Dualizing (2.18), we obtain

\[ 0 \to \mathcal{O}(0,-1) \to \Omega(1) \boxtimes_X \mathcal{O} \to \mathcal{E}^{{\vee}} \to 0. \]

Taking its $(k+i)$th exterior power and twisting by $\mathcal {O}(0,i-1)$, we obtain

\[ 0 \to \Lambda^{k+i-1}\mathcal{E}^{{\vee}}(0,i-2) \to \Omega^{k+i}(k+i) \boxtimes_X \mathcal{O}(i-1) \to \Lambda^{k+i}\mathcal{E}^{{\vee}}(0,i-1) \to 0. \]

Using (2.19), we can rewrite this as

(2.20)\begin{equation} 0 \to \Lambda^{k-i+1}\mathcal{E}({-}1,i-1) \to \Omega^{k+i}(k+i) \boxtimes_X \mathcal{O}(i-1) \to \Lambda^{k-i}\mathcal{E}({-}1,i) \to 0. \end{equation}

Now we interpret these sequences as mutations of (2.10).

• – For $i = k$, the second arrow in (2.20) is the unique (by (2.11)) morphism

  \[ \Omega^{2k}(2k) \boxtimes_X \mathcal{O}(k-1) \to \mathcal{O}({-}1,k), \]
  therefore its kernel (up to shift) is the right mutation of $\Omega ^{2k}(2k) \boxtimes _X \mathcal {O}(k-1)$ through the line bundle $\mathcal {O}(-1,k)$, and so by (2.20) the result of the mutation is $\mathcal {E}(-1,k-1)$.

• – For $i = k-1$, the second arrow in (2.20) is the unique morphism

  \[ \Omega^{2k-1}(2k-1) \boxtimes_X \mathcal{O}(k-2) \to \mathcal{E}({-}1,k-1), \]
  therefore its kernel (up to shift) is the right mutation of $\Omega ^{2k-1}(2k-1) \boxtimes _X \mathcal {O}(k-2)$ through the bundle $\mathcal {E}(-1,k-1)$, and so the result of the mutation is $\Lambda ^{2}\mathcal {E}(-1,k-2)$. Note here that the intermediate mutation of $\Omega ^{2k-1}(2k-1) \boxtimes _X \mathcal {O}(k-2)$ through $\mathcal {O}(-1,k)$ is trivial, because these objects are completely orthogonal by (2.11).

Continuing in the same manner, we conclude that the mutation of the vector bundle $\Omega ^{2k-j}(2k-j) \boxtimes _X \mathcal {O}(k-j-1)$ through

\[ \langle \Omega ^{2k-j+1}(2k-j+1) \boxtimes _X \mathcal {O}(k-j),\ldots , \Omega ^{2k}(2k) \boxtimes _X \mathcal {O}(k-1), \mathcal {O}(-1,k) \rangle \]

is the same as its mutation through $\langle \mathcal {O}(-1,k), \mathcal {E}(-1,k-1),\ldots , \Lambda ^{j}\mathcal {E}(-1,k-j) \rangle$, and is the same as its mutation through the last object $\Lambda ^{j}\mathcal {E}(-1,k-j)$. Therefore, it is realized by the complex (2.20) with $i = k - j$, hence the result of the mutation is $\Lambda ^{j+1}\mathcal {E}(-1,k-j-1)$.

Thus, we see that the category $\mathcal {R}$ is generated by the exceptional collection

(2.21)\begin{align} \mathcal{R} = \langle \mathcal{O}({-}1,k), &\mathcal{E}({-}1,k-1),\ldots, \Lambda^{k-1}\mathcal{E}({-}1,1), \Lambda^{k}\mathcal{E}({-}1,0),\nonumber\\ &\quad \mathcal{O} \boxtimes_X \Omega^{k+1}(k+1), \mathcal{O}(1) \boxtimes_X \Omega^{k+2}(k+2),\ldots, \mathcal{O}(k-1) \boxtimes_X \Omega^{2k}(2k) \rangle, \end{align}

and moreover, the only $\operatorname {Ext}$-spaces between the objects in the first row and the second row of (2.21) are

(2.22)\begin{equation} \operatorname{Ext}^{{\bullet}}(\Lambda^{i}\mathcal{E}({-}1,k-i),\mathcal{O}(j-1) \boxtimes_X \Omega^{k+j}(k+j)) = \begin{cases} \Bbbk[i-k], & \text{if } j = 1\\ 0, & \text{if } 2 \le j \le k. \end{cases} \end{equation}

Note also that by (2.19) the last term in the first line of (2.21) is isomorphic to $\Lambda ^{k}\mathcal {E}^{\vee }(0,-1)$.

For the second half of mutations, we note that, applying an outer automorphism of $X$ to (2.20), we obtain exact sequences

(2.23)\begin{equation} 0 \to \Lambda^{k-i+1}\mathcal{E}^{{\vee}}(i-1,-1) \to \mathcal{O}(i-1) \boxtimes_X \Omega^{k+i}(k+i) \to \Lambda^{k-i}\mathcal{E}^{{\vee}}(i,-1) \to 0. \end{equation}

Now we interpret these sequences as mutations. Note that $\Lambda ^{k}\mathcal {E}(-1,0) \cong \Lambda ^{k}\mathcal {E}^{\vee }(0,-1)$ which follows from (2.19).

• – For $i = 1$, the first arrow in (2.23) is the unique (by (2.22)) morphism

  \[ \Lambda^{k}\mathcal{E}^{{\vee}}(0,-1) \to \mathcal{O} \boxtimes_X \Omega^{k+1}(k+1), \]
  therefore its cokernel (up to shift) is the left mutation of the object $\mathcal {O} \boxtimes _X \Omega ^{k+1}(k+1)$ through the bundle $\Lambda ^{k}\mathcal {E}^{\vee }(0,-1)$, and so the result of the mutation is $\Lambda ^{k-1}\mathcal {E}^{\vee }(1,-1)$.

• – For $i = 2$, the first arrow in (2.23) is the unique morphism

  \[ \Lambda^{k-1}\mathcal{E}^{{\vee}}(1,-1) \to \mathcal{O}(1) \boxtimes_X \Omega^{k+2}(k+2), \]
  therefore its cokernel (up to shift) is the left mutation of $\mathcal {O}(1) \boxtimes _X \Omega ^{k+2}(k+2)$ through the bundle $\Lambda ^{k-1}\mathcal {E}^{\vee }(1,-1)$, and so the result of the mutation is $\Lambda ^{k-2}\mathcal {E}^{\vee }(2,-1)$. Note here that the intermediate mutation of $\mathcal {O}(1) \boxtimes _X \Omega ^{k+2}(k+2)$ through $\Lambda ^{k}\mathcal {E}^{\vee }(0,-1)$ is trivial, because these objects are completely orthogonal by (2.22).

Continuing in the same manner, we conclude that the mutation of $\mathcal {O}(i-1) \boxtimes _X \Omega ^{k+i}(k+i)$ through $\langle \Lambda ^{k}\mathcal {E}^{\vee }(0,-1), \mathcal {O} \boxtimes _X \Omega ^{k+1}(k+1),\ldots , \mathcal {O}(i-2) \boxtimes _X \Omega ^{k+i-1}(k+i-1) \rangle$ is the same as its mutation through $\langle \Lambda ^{k-i+1}\mathcal {E}^{\vee }(i-1,-1),\ldots , \Lambda ^{k-1}\mathcal {E}^{\vee }(1,-1), \Lambda ^{k}\mathcal {E}^{\vee }(0,-1) \rangle$, and the same as its mutation through the first object $\Lambda ^{k-i+1}\mathcal {E}^{\vee }(i-1,-1)$. Therefore, it is realized by the exact sequence (2.23), hence the result of the mutation is $\Lambda ^{k-i+1}\mathcal {E}^{\vee }(i-1,-1)$.

Thus, we see that the category $\mathcal {R}$ is generated by (2.4). Moreover, we see that the only $\operatorname {Ext}$-spaces between the objects in (2.4) are

\begin{align*} &\operatorname{Ext}^{j-i}(\Lambda^{i}\mathcal{E}\hphantom{{}^{{\vee}}}(0,k-i), \Lambda^{j}\mathcal{E}(0,k-j)) = \Bbbk, \quad\ \,\, 0 \le i \le j \le k,\\ &\operatorname{Ext}^{j-i}(\Lambda^{i}\mathcal{E}^{{\vee}}(0,k-i), \Lambda^{j}\mathcal{E}^{{\vee}}(0,k-j)) = \Bbbk, \quad 0 \le i \le j \le k. \end{align*}

In other words, the configuration of these exceptional objects is the following:

So, this exceptional collection corresponds to shifts of projective modules in a quiver of Dynkin type $\mathrm {A}_{2k+1}$.

Theorem 3.1 There exists a full exceptional collection

(3.9)\begin{align} {{\mathbf{D}^{\mathrm{b}}}}(\operatorname{OG}(2,2n)) &= \langle \mathcal{U}^{2\omega_{n-1}}({-}1), \mathcal{U}^{2\omega_{n}}({-}1) , \mathcal{A} , \nonumber\\ &\qquad \mathcal{B}(1) ,\ldots , \mathcal{B}(n-2) , \mathcal{A}(n-1) ,\ldots ,\mathcal{A}(2n-4) \rangle. \end{align}

In particular, the subcategory $\mathcal {A}$ extends to an $\operatorname {Aut}(\operatorname {OG}(2,2n))$-invariant rectangular Lefschetz collection

(3.10)\begin{equation} {{\mathbf{D}^{\mathrm{b}}}}(\operatorname{OG}(2,2n)) = \langle \mathcal{R}, \mathcal{A}, \mathcal{A}(1),\ldots, \mathcal{A}(2n - 4) \rangle. \end{equation}

Moreover, the residual category $\mathcal {R}$ is generated by an $\operatorname {Aut}(\operatorname {OG}(2,2n))$-invariant exceptional collection and is equivalent to the derived category of representations of the Dynkin quiver $\mathrm {D}_n$.

Theorem 3.7 (Borel–Bott–Weil) Let $\lambda \in \mathbf {P}_\mathrm {L}^{+}$. If the weight $\lambda + \rho$ lies on a wall of a Weyl chamber for the $\mathrm {W}$-action, then

\[ H^{{\bullet}}(\mathrm{G}/\mathrm{P},\mathcal{U}^{\lambda}) = 0. \]

Otherwise, if $w \in \mathrm {W}$ is the unique element such that the weight $w(\lambda + \rho )$ is dominant, then

\[ H^{{\bullet}}(\mathrm{G}/\mathrm{P},\mathcal{U}^{\lambda}) = \mathrm{V}_\mathrm{G}^{w(\lambda + \rho) - \rho}[-\ell(w)]. \]

Proof. By (3.3) we have $a\omega _1 = (a,0,0,\ldots ,0)$. By Lemma 3.5 and (3.12) the weight corresponding to $\big ( \mathcal {U}^{a \omega _1} \big )^{\vee }$ is $(0,-a,0,\ldots ,0)$; by (3.3) this is equal to $a\omega _1 - a\omega _2$, hence the claim. The other results are proved analogously.

Proof. First, we note that the bundles $\mathcal {U}^{k\omega _1}$ and $\mathcal {U}^{l\omega _1}$ correspond to representations of $\mathrm {L}$ pulled back from representations $\mathrm {V}_{\operatorname {GL}_2}^{k,0}$ and $\mathrm {V}_{\operatorname {GL}_2}^{l,0}$ via the map (3.4), so (3.14) follows from the Clebsch–Gordan rule

\[ \mathrm{V}_{\operatorname{GL}_2}^{k_1,k_2} \otimes \mathrm{V}_{\operatorname{GL}_2}^{l_1,l_2} = \bigoplus_{j = 0}^{\min\{k_1 - k_2,l_1 - l_2\}} \mathrm{V}_{\operatorname{GL}_2}^{k_1 + l_1 - j, k_2 + l_2 + j}. \]

For (3.15) and (3.16) we need a slightly more detailed understanding of the Levi group $\mathrm {L}$. Let $\operatorname {{\widetilde {GL}}}(2) \to \operatorname {GL}(2)$ be the unique connected double covering. Then there is a double covering

(3.17)\begin{equation} \operatorname{{\widetilde{GL}}}(2) \times \operatorname{Spin}(2(n-2)) \to \mathrm{L}. \end{equation}

To understand the tensor product of representations of $\mathrm {L}$ it is enough to understand the tensor product of their pullbacks via the map (3.17).

We will denote by $\omega '_1$, $\omega '_2$ the fundamental weights of $\operatorname {{\widetilde {GL}}}(2)$ and by $\omega '_i$, $3 \le i \le n$, the fundamental weights of $\operatorname {Spin}(2(n-2))$. The double covering (3.17) induces an embedding of weight lattices

\[ \mathbf{P}_\mathrm{L} \hookrightarrow \mathbf{P}_{\operatorname{{\widetilde{GL}}}(2)} \oplus \mathbf{P}_{\operatorname{Spin}(2(n-2))} \]

and identifies the corresponding ${{\mathbb {Q}}}$-vector spaces. Therefore, the weights $\omega '_i$ can be thought of as rational weights for $\mathrm {L}$, that is, as rational weights of $\operatorname {Spin}(2n)$. It is easy to see that they can be expressed in the basis of fundamental weights $\omega _i$ as follows:

(3.18)\begin{equation} \omega'_1 = \omega_1,\quad \omega'_2 = \tfrac 12\omega_2,\quad \omega'_i =\begin{cases} \omega_i - \omega_2, & \text{if } 3 \le i \le n-2,\\ \omega_i - \tfrac 12\omega_2, & \text{if } i \in \{n-1,n\}. \end{cases} \end{equation}

By (3.18) the representation $\mathrm {V}_\mathrm {L}^{k\omega _1}$ pulls back to $\mathrm {V}_{\operatorname {{\widetilde {GL}}}_2}^{k,0}$, while $\mathrm {V}_\mathrm {L}^{\lambda }$ with $\lambda = \sum _{i=2}^{n} \lambda _i \omega _i$ pulls back to $\mathrm {V}_{\operatorname {{\widetilde {GL}}}(2)}^{c(\lambda ),c(\lambda )} \otimes \mathrm {V}_{\operatorname {Spin}(2(n-2))}^{\lambda '}$, where $\lambda ' = \sum _{i=3}^{n} \lambda _i \omega '_i$ and $c(\lambda ) = 2\sum _{i=2}^{n-2}\lambda _i + \lambda _{n-1} + \lambda _n$. Clearly,

\[ \mathrm{V}_{\operatorname{{\widetilde{GL}}}_2}^{k,0} \otimes \Big( \mathrm{V}_{\operatorname{{\widetilde{GL}}}(2)}^{c(\lambda),c(\lambda)} \otimes V_{\operatorname{Spin}(2(n-2))}^{\lambda'} \Big) \cong \mathrm{V}_{\operatorname{{\widetilde{GL}}}(2)}^{k + c(\lambda),c(\lambda)} \otimes \mathrm{V}_{\operatorname{Spin}(2(n-2))}^{\lambda'}, \]

and the right-hand side corresponds to $\mathrm {V}_\mathrm {L}^{k\omega _1 + \lambda }$. This proves (3.15).

Similarly, $\mathrm {V}_\mathrm {L}^{\omega _3 - \omega _2}$ corresponds to $\mathrm {V}_{\operatorname {Spin}(2(n-2))}^{\omega '_3}$ (which, according to our conventions, is the first fundamental representation of $\operatorname {Spin}(2(n-2))$). By [Reference Vinberg and OnishchikVO95, Table 5] its middle exterior power is isomorphic to the direct sum of two irreducible representations with highest weights $2\omega '_n$ and $2\omega '_{n-1}$. By (3.18) these weights correspond to the weights $2\omega _n - \omega _2$ and $2\omega _{n-1} - \omega _2$, and the claim follows. This proves (3.16).

Proof. These are special cases of [Reference Kuznetsov and PolishchukKP16, Lemma 5.2]; we provide a proof for completeness.

In view of (3.13), if (3.19) holds, then all coordinates of $\lambda + \rho$ have absolute values less than $n - 1$. Therefore, at least two of them have the same absolute values. Similarly, if (3.20) holds, then all but one of the coordinates of $\lambda + \rho$ have absolute values less than $n - 2$. Therefore, again at least two of them have the same absolute values. In both cases we see that $\lambda + \rho$ lies on one of the walls (3.11), and by the Borel–Bott–Weil theorem we conclude that $\mathcal {U}^{\lambda }$ is acyclic.

Proof. We will show that $\operatorname {Ext}^{\bullet }(S^{k}\mathcal {U}^{\vee }(t),S^{l}\mathcal {U}^{\vee })$ is zero for $0 \le k,l \le n-2$ and $0 \le t \le 2n-4$, except for $t = 0$, $k \le l$ and $t \in \{n - 2, n - 1\}$, $k = l = n-2$. By (3.6) and Serre duality we can assume $0 \le t \le n - 2$.

By Corollary 3.8 and (3.14) we have

\begin{align*} (S^{k}\mathcal{U}^{{\vee}}(t))^{{\vee}} \otimes S^{l}\mathcal{U}^{{\vee}} &\cong S^{k}\mathcal{U}^{{\vee}}({-}k-t) \otimes S^{l}\mathcal{U}^{{\vee}} \\ &\cong \bigoplus_{j = 0}^{\min\{k,l\}} S^{k + l - 2j}\mathcal{U}^{{\vee}}(j - k - t) = \bigoplus_{j = 0}^{\min\{k,l\}} \mathcal{U}^{(l - j - t,j - k - t,0,\ldots,0)}. \end{align*}

So, it is enough to compute the cohomology of bundles $\mathcal {U}^{\lambda }$ with $\lambda = (l - j - t,j - k - t,0,\ldots ,0)$, so that

\[ 2 - n \le \lambda_1 \le n - 2,\quad 4 - 2n \le \lambda_2 \le 0,\quad \lambda_3 = \cdots = \lambda_n = 0. \]

It is convenient to distinguish three cases.

Case 1. If $4 - 2n < \lambda _2 < 0$, then $|\lambda _i + n - i| < n - 2$ for all $i \ge 2$. Hence, (3.20) holds and $\mathcal {U}^{\lambda }$ is acyclic by Lemma 3.10.

Case 2. If $\lambda _2 = 0$, then we necessarily have $t = 0$ and $j = k$. Further, since $j \le l$, we obtain $k \le l$. In particular, the weight $\lambda$ is dominant, hence $H^{>0}(\operatorname {OG}(2,2n),\mathcal {U}^{\lambda }) = 0$. Moreover, if $k = l$, then $\lambda = 0$ and in this case the cohomology of $\mathcal {U}^{\lambda }$ is isomorphic to $\Bbbk [0]$.

A combination of cases 1 and 2 proves that $\mathcal {O}, \mathcal {U}^{\vee },\ldots , S^{n-2}\mathcal {U}^{\vee }$ is a strong exceptional collection.

Case 3. If $\lambda _2 = 4 - 2n$, then we necessarily have $k = t = n - 2$ and $j = 0$. In this case we have inequalities $2 - n \le \lambda _1 \le 0$.

If $\lambda _1 < 0$, then $|\lambda _i + n - i| < n - 1$ for all $i$. Hence (3.19) holds and $\mathcal {U}^{\lambda }$ is acyclic by Lemma 3.10.

If $\lambda _1 = 0$, then $l = n - 2$, and we have $\mathcal {U}^{\lambda } = \mathcal {U}^{(0,4-2n,0,\ldots ,0)}$. In this case the Borel–Bott–Weil theorem implies that the cohomology of $\mathcal {U}^{\lambda }$ is isomorphic to $\Bbbk [4 - 2n]$. Indeed, we have $\lambda + \rho = (n-1,2-n,n-3,\ldots ,0) = w\rho$, where

\[ w = (\mathrm{s}_2 \cdot \ldots \cdot \mathrm{s}_{n-2}) \cdot \mathrm{s}_{n-1} \cdot \mathrm{s}_n \cdot (\mathrm{s}_{n-2} \cdot \ldots \cdot \mathrm{s}_2) \]

and $\mathrm {s}_i$ are the simple reflections. This completes the proof of the lemma.

Proof. We will compute $\operatorname {Ext}^{\bullet }(\mathcal {U}^{a\omega _i}(t),\mathcal {U}^{b\omega _j})$ where $i,j \in \{n - 1,n\}$, and

• – either $a,b \in \{1,2\}$ and $(a,b) \ne (2,2)$, and $1 \le t \le 2n - 4$,

• – or $a,b \in \{1,2\}$ and $(a,b) \ne (1,2)$, and $t = 0$.

By (3.6) and Serre duality we can assume $0 \le t \le n - 2$.

We take $\lambda = a\omega _{i} + t\omega _2 = (t+ a/2,t+ a/2, a/2,\ldots , a/2,\pm a/2)$ and $\mu = b\omega _{j} = ( b/2, b/2, b/2, \ldots\,$, $ b/2,\pm b/2)$; the signs of the last coordinate depend on whether $i$ and $j$ are $n-1$ or $n$. Since the $\mathrm {W}_\mathrm {L}$-orbit of $\lambda$ consists of points $(t + a/2,t + a/2,\pm a/2,\ldots ,\pm a/2,\pm a/2)$ (with the parity of the number of negative signs depending on $i$), by Lemma 3.6 we have

(3.21)\begin{equation} (\mathcal{U}^{\lambda})^{{\vee}} \otimes \mathcal{U}^{\mu} = \bigoplus_{\nu} (\mathcal{U}^{\nu})^{{\oplus} m(\lambda,\mu,\nu)}, \end{equation}

where $\nu$ runs over the integral or half-integral points satisfying

\[ \nu_1 = \nu_2 = \frac{b-a}2 - t, \quad \nu_3,\ldots,\nu_{n-1} = \frac{b \pm a}2 \in \bigg[-\frac 12,2\bigg], \quad \nu_n = \frac{\pm b \pm a}2 \in [{-}2,2]. \]

(In fact we could further restrict the combinations of signs that appear in $\nu$, but since we are interested in an upper bound on direct summands of $(\mathcal {U}^{\lambda })^{\vee } \otimes \mathcal {U}^{\mu }$, we can neglect this.) It is convenient to distinguish three cases.

Case 1. If $1 \le t \le n-2$ and $(a,b) \ne (2,2)$, then

\[ \nu_1 = \nu_2 \in \big[\tfrac 32 - n, -\tfrac 12\big], \quad \nu_3,\ldots,\nu_{n-1},\nu_n \in \big[-\tfrac 32,\tfrac 32\big], \]

For any such point we have $|\nu _i + n - i| < n - 1$ for all $i$, hence (3.19) holds for $\nu$ and, therefore, all summands in (3.21) are acyclic by Lemma 3.10.

Case 2. If $t = 0$ and $a > b$, then

\[ \nu_1 = \nu_2 ={-}\tfrac 12, \quad \nu_3,\ldots,\nu_{n-1} \in \big[-\tfrac 12,\tfrac 32\big], \quad \nu_n \in \big[-\tfrac 32, \tfrac 32\big]. \]

For any such point we still have $|\nu _i + n - i| < n - 1$ for all $i$, hence again all summands in (3.21) are acyclic.

Case 3. If $t = 0$ and $a = b$, then

\[ \nu_1 = \nu_2 = 0, \quad \nu_3,\ldots,\nu_{n-1} \in [0,2], \quad \nu_n \in [{-}2, 2]. \]

It is easy to see that the only non-acyclic bundle among $\mathcal {U}^{\nu }$ is the trivial bundle. Indeed, if $\tilde \nu = \nu + \rho$ then $(\tilde \nu _1, \tilde \nu _2) = (n-1,n-2)$ and $n - 1 \ge \tilde \nu _3 \ge \cdots \ge \tilde \nu _{n-1} \ge 1$, and the only possibility to satisfy these restriction and avoid the walls (3.11) is to have $\tilde \nu _i = n - i$ for $1 \le i \le n - 1$. Finally, $0 \le \tilde \nu _n \le 2$, and since $n \ge 3$, it must be zero to avoid the walls. By Lemma 3.6 the trivial bundle is a summand of $(\mathcal {U}^{\lambda })^{\vee } \otimes \mathcal {U}^{\mu }$ only when $\lambda = \mu$ and then its multiplicity is equal to 1. Hence, in this case we have $\operatorname {Ext}^{\bullet }(\mathcal {U}^{\lambda },\mathcal {U}^{\mu }) = \Bbbk [0]$. This completes the proof of the lemma.

Proof. By Corollary 3.8 and (3.15) we have

\[ (\mathcal{U}^{\omega_j}(t))^{{\vee}} \otimes S^{k}\mathcal{U}^{{\vee}} \cong \mathcal{U}^{\omega_{j'} - (t+1)\omega_2} \otimes \mathcal{U}^{k \omega_1} \cong \mathcal{U}^{k\omega_1 - (t+1)\omega_2 + \omega_{j'}}, \]

where $j' = j$ if $n$ is even and $j' = 2n - 1 - j$ if $n$ is odd. So, it is enough to compute the cohomology of bundles $\mathcal {U}^{\lambda }$ with

\[ \lambda = \big(k - t - \tfrac 12, - t - \tfrac 12, \tfrac 12,\ldots,\tfrac 12,\pm\tfrac 12 \big). \]

If $0 \le t \le 2n - 5$, then $|\lambda _i + n - i| < n - 2$ for $i \ge 2$. Hence (3.20) holds and the bundle $\mathcal {U}^{\lambda }$ is acyclic by Lemma 3.10. Similarly, if $t = 2n - 4$, then we have $|\lambda _i + n - i| < n - 1$ for all $i$. Hence condition (3.19) holds and the bundle $\mathcal {U}^{\lambda }$ is acyclic by Lemma 3.10.

Proof. As before, we have

\[ (S^{k}\mathcal{U}^{{\vee}}(t))^{{\vee}} \otimes \mathcal{U}^{2\omega_j} \cong S^{k}\mathcal{U}^{{\vee}}({-}k-t) \otimes \mathcal{U}^{2\omega_j} \cong \mathcal{U}^{k \omega_1 - (k + t)\omega_2 + 2\omega_j}. \]

So, it is enough to compute the cohomology of the bundle $\mathcal {U}^{\lambda }$ with

\[ \lambda = (1-t,1-k-t,1,\ldots,1,\pm 1). \]

For this we directly apply the Borel–Bott–Weil theorem. We have

\[ \lambda + \rho = (n - t, n - 1 - k - t, n - 2,\ldots, 2, \pm 1). \]

It is convenient to distinguish several cases.

Case 1. If $2 \le t \le 2n - 2$ and $t \ne n$, then the absolute value of the first coordinate of $\lambda + \rho$ is equal to the absolute value of one of the last $n - 2$ coordinates. Hence, the bundle $\mathcal {U}^{\lambda }$ is acyclic.

Case 2. If $t = n$ and $0 \le k \le n - 3$, then $n - 1 - k - t = - k - 1$, and the absolute value of the second coordinate of $\lambda + \rho$ is equal to the absolute value of one of the last $n - 2$ coordinates. Hence, $\mathcal {U}^{\lambda }$ is acyclic.

Case 3. If $t = n$ and $k = n - 2$, then $\lambda + \rho = (0, 1 - n, n - 2,\ldots , 2, \pm 1) = w\rho$, where

\[ w = (\mathrm{s}_{n-1} \cdot \mathrm{s}_{n-2} \cdot \ldots \cdot \mathrm{s}_2) \cdot \mathrm{s}_1 \cdot (\mathrm{s}_2 \cdot \ldots \cdot \mathrm{s}_{n-2}) \cdot \mathrm{s}_{n-1} \cdot \mathrm{s}_n \cdot (\mathrm{s}_{n-2} \cdot \ldots \cdot \mathrm{s}_2) \]

and $\mathrm {s}_i$ are the simple reflections. By the Borel–Bott–Weil theorem the cohomology of the bundle $\mathcal {U}^{\lambda }$ equals $\Bbbk [5 - 3n]$.

Case 4. If $t = 1$ and $0 \le k \le n - 3$, then $n - 1 - k - t = n - 2 - k$, and the absolute value of the second coordinate of $\lambda + \rho$ is equal to the absolute value of one of the last $n - 2$ coordinates. Hence, $\mathcal {U}^{\lambda }$ is acyclic.

Case 5. If $t = 1$ and $k = n - 2$, then $\lambda + \rho = (n - 1, 0, n - 2,\ldots , 2, \pm 1) = w\rho$, where

\[ w = \mathrm{s}_{n-1} \cdot \mathrm{s}_{n-2} \cdot \ldots \cdot \mathrm{s}_2 \]

and $\mathrm {s}_i$ are the simple reflections. By the Borel–Bott–Weil theorem the cohomology of the bundle $\mathcal {U}^{\lambda }$ equals $\Bbbk [2 - n]$. This completes the proof of the lemma.

Proof. A combination of Lemmas 3.11–3.14, and Serre duality proves the proposition.

Proof. If $n$ is even, we apply [Reference KuznetsovKuz08, Proposition 6.7] and conclude that there is a filtration on the bundle $\mathbf {S}_+ \otimes \mathcal {S}_+^{\vee } \oplus \mathbf {S}_- \otimes \mathcal {S}_-^{\vee }$ with factors of the form $\Lambda ^{2s}\mathcal {U}^{\perp }$, where $0 \le s \le n - 1$. Tensoring by $\mathcal {O}(1)$ and taking into account the isomorphisms

\[ \mathcal{S}_+^{{\vee}}(1) \cong \mathcal{S}_+, \quad \mathcal{S}_-^{{\vee}}(1) \cong \mathcal{S}_- \]

(Corollary 3.8), we obtain a filtration on $\mathbf {S}_+ \otimes \mathcal {S}_+ \oplus \mathbf {S}_- \otimes \mathcal {S}_-$ with factors of the form $\Lambda ^{2s}\mathcal {U}^{\perp }(1)$, where $0 \le s \le n - 1$.

Similarly, if $n$ is odd, the argument of [Reference KuznetsovKuz08, Proposition 6.7] proves that there is a filtration on $\mathbf {S}_+ \otimes \mathcal {S}_-^{\vee } \oplus \mathbf {S}_- \otimes \mathcal {S}_+^{\vee }$ with factors of the form $\Lambda ^{2s+1}\mathcal {U}^{\perp }$, where $0 \le s \le n - 2$. Tensoring by $\mathcal {O}(1)$ and taking into account isomorphisms

\[ \mathcal{S}_+^{{\vee}}(1) \cong \mathcal{S}_-, \quad \mathcal{S}_-^{{\vee}}(1) \cong \mathcal{S}_+, \]

we finally obtain a filtration on $\mathbf {S}_+ \otimes \mathcal {S}_+ \oplus \mathbf {S}_- \otimes \mathcal {S}_-$ with factors of the form $\Lambda ^{2s+1}\mathcal {U}^{\perp }(1)$, where we have $0 \le s \le n - 2$.

Thus, in both cases we have a filtration on $\mathbf {S}_+ \otimes \mathcal {S}_+ \oplus \mathbf {S}_- \otimes \mathcal {S}_-$ with factors

\[ \Lambda^{2n - 2 - r}\mathcal{U}^{{\perp}}(1), \Lambda^{2n - 4 - r}\mathcal{U}^{{\perp}}(1),\ldots, \Lambda^{n}\mathcal{U}^{{\perp}}(1), \Lambda^{n - 2}\mathcal{U}^{{\perp}}(1),\ldots, \Lambda^{2 + r}\mathcal{U}^{{\perp}}(1), \Lambda^{r}\mathcal{U}^{{\perp}}(1), \]

where $r = 0$ if $n$ is even and $r = 1$ if $n$ is odd. Note that the total number of factors $n - r$ is even in both cases. This means that there is an exact sequence of vector bundles

\[ 0 \to F_- \to \mathbf{S}_+\,{\otimes}\, \mathcal{S}_+\,{\oplus}\, \mathbf{S}_-\,{\otimes}\, \mathcal{S}_- \to F_+ \to 0, \]

where $F_-$ has a filtration with factors $\Lambda ^{2n - 2 - r}\mathcal {U}^{\perp }(1), \Lambda ^{2n - 4 - r}\mathcal {U}^{\perp }(1),\ldots , \Lambda ^{n}\mathcal {U}^{\perp }(1)$ and $F_+$ has a filtration with factors $\Lambda ^{n - 2}\mathcal {U}^{\perp }(1),\ldots , \Lambda ^{2 + r}\mathcal {U}^{\perp }(1), \Lambda ^{r}\mathcal {U}^{\perp }(1)$. So, it is enough to show that $F_-$ has a left resolution given by the first half of (3.25), and that $F_+$ has a right resolution given by the second half of (3.25).

Indeed, each factor of the filtration of $F_-$ can be rewritten as

\[ \Lambda^{n + 2k}\mathcal{U}^{{\perp}}(1) \cong \Lambda^{n - 2 - 2k}(\mathcal{V}/\mathcal{U}), \]

where $0 \le k \le (n - 2 - r)/2$, hence has the resolution (3.22) with $m = n - 2 - 2k$. Since by Lemma 3.11 the exceptional collection $\{ S^{i}\mathcal {U} \}_{i = 0}^{n-2}$ is strong, the extensions of the filtration factors $\Lambda ^{n + 2k}\mathcal {U}^{\perp }(1)$ in $F_-$ can be realized by a bicomplex with rows given by (3.22) with $m = n - 2 - 2k$ shifted by $-k$. The totalization of this bicomplex can be written as

\[ S^{n-2} \mathcal{U} \to W_1 \otimes S^{n-3} \mathcal{U} \to W_2 \otimes S^{n-4} \mathcal{U} \to \cdots \to W_{n-2} \otimes \mathcal{O} \]

and thus provides a left resolution for $F_-$.

Similarly, each factor $\Lambda ^{n-2-2k}\mathcal {U}^{\perp }(1)$ of the filtration of $F_+$ has resolution (3.23) twisted by $\mathcal {O}(1)$ (where $m = n - 2 - 2k$), and the above argument shows that $F_+$ has the right resolution

\[ W_{n-2} \otimes \mathcal{O}(1) \to \cdots \to W_2 \otimes S^{n-4} \mathcal{U}^{{\vee}}(1) \to W_1 \otimes S^{n-3} \mathcal{U}^{{\vee}}(1) \to S^{n-2} \mathcal{U}^{{\vee}}(1). \]

This completes the proof of the proposition.

Proof. The bicomplex is just the $m$th exterior power of the complex

\[ \mathcal{U} \to \mathcal{V} \to \mathcal{U}^{{\vee}}, \]

which is quasi-isomorphic to $\mathcal {U}^{\perp }/\mathcal {U}$, hence the claim (cf. the argument of [Reference KuznetsovKuz08, Lemma 7.3]).

Proof. The inclusion (3.29) follows immediately from comparison of (3.28), (3.8) and (3.27). Thus, we concentrate here on the inclusion (3.30).

We split objects $E \in \mathbf {E}'$ into two classes:

• – $\mathcal {S}_+(t)$ with $t \in [2-n,n-3]$ (spinor bundles), and

• – $S^{k} \mathcal {U}^{\vee }(t)$ with $k \in [0,n-3]$ and $t \in [2-n,n-3]$ (tautological bundles),

and treat these separately.

The class of spinor bundles is easy: from Lemma 3.18 we conclude

\[ \mathcal{S}_+(i) \otimes \mathcal{U}^{{\vee}} \in \langle \mathcal{S}_-(i), \mathcal{O}(i+1), \mathcal{S}_-(i+1) \rangle, \]

and the inclusion (3.30) for $E = \mathcal {S}_+(i)$ follows.

The class of tautological bundles is a bit more tedious. First, note that by (3.14) we have

\[ S^{k} \mathcal{U}^{{\vee}} \otimes \mathcal{U}^{{\vee}} = S^{k+1} \mathcal{U}^{{\vee}} \oplus S^{k-1}\mathcal{U}^{{\vee}}(1). \]

Thus, the inclusion (3.30) would follow from the inclusions

(3.31)\begin{equation} S^{j}\mathcal{U}^{{\vee}}(t) \in \langle \mathbf{E} \rangle, \quad 0 \le j \le n -2,\quad 2 - n \le t \le n - 2. \end{equation}

So, by proving these inclusions, we will prove the lemma.

For $j \le n - 3$ and for $j = n - 2$, $1 \le t \le n - 2$, the inclusions (3.31) follow immediately from (3.27). Furthermore, using (3.25) twisted by $\mathcal {O}(n - 2 + t)$ and isomorphisms

\[ S^{l} \mathcal{U} \simeq S^{l} \mathcal{U}^{{\vee}}({-}l), \]

we deduce (3.31) for $j = n - 2$ and $2 - n \le t \le -1$.

So, it only remains to show that

\[ S^{n-2}\mathcal{U}^{{\vee}} \in \langle \mathbf{E} \rangle. \]

For this we consider the double complex of Lemma 3.17 with $m = n-2$. Using the direct sum decompositions

\[ S^{k}\mathcal{U}^{{\vee}} \otimes S^{l}\mathcal{U} \cong \bigoplus_{j = 0}^{\min\{k,l\}} S^{k + l - 2j}\mathcal{U}^{{\vee}}(j - l) \]

and the cases of the inclusion (3.31) proved above, we see that all the terms of the double complex with a possible exception for the rightmost term $S^{n-2}\mathcal {U}^{\vee }$ are contained in the subcategory generated by $\mathbf {E}$. By (3.16) and (3.26) the unique cohomology sheaf

\[ \mathcal{U}^{2\omega_{n-1}}({-}1) \oplus \mathcal{U}^{2\omega_n}({-}1) \]

of the double complex is contained in $\langle \mathbf {E} \rangle$, therefore the last term $S^{n-2}\mathcal {U}^{\vee }$ is in the subcategory generated by $\mathbf {E}$.

Proof. The assumption of the proposition can be rewritten as

(3.33)\begin{equation} \operatorname{Ext}^{{\bullet}}(E,F) = H^{{\bullet}}(\operatorname{OG}(2,V),E^{{\vee}} \otimes F)=0 \quad \forall E \in \mathbf{E}. \end{equation}

Now take any non-isotropic vector $v$ and any bundle $E \in \mathbf {E}'$, and tensor (3.32) by $E^{\vee } \otimes F$. We obtain an exact sequence

(3.34)\begin{equation} 0 \to \mathcal{O}({-}1) \otimes E^{{\vee}} \otimes F \to \mathcal{U} \otimes E^{{\vee}} \otimes F \to E^{{\vee}} \otimes F \to i_{v*} i_v^{*} (E^{{\vee}} \otimes F) \to 0. \end{equation}

From Lemma 3.19 and (3.33) it follows that the cohomology groups of the first three terms of this complex vanish. Hence, also the cohomology of $i_{v*} i_v^{*} (E^{\vee } \otimes F)$ has to vanish, and we conclude that

(3.35)\begin{equation} H^{{\bullet}} (\operatorname{OG}(2,V), i_{v*} i_v^{*} (E^{{\vee}} \otimes F)) = H^{{\bullet}} (\operatorname{OG}(2,v^{{\perp}}), i_v^{*} (E^{{\vee}} \otimes F)) = \operatorname{Ext}^{{\bullet}}(i_v^{*} E, i_v^{*} F) = 0 \end{equation}

for any $E \in \mathbf {E}'$. In other words, the object $i_v^{*}F$ is orthogonal to all objects $i_v^{*}E$ for $E \in \mathbf {E}'$.

Set

\[ \mathcal{C} := \langle \mathcal{O}, i_v^{*}(\mathcal{U}^{{\vee}}),\ldots, i_v^{*}(S^{n-3}\mathcal{U}^{{\vee}}), i_v^{*}(\mathcal{S}_+) \rangle \subset {{\mathbf{D}^{\mathrm{b}}}}(\operatorname{OG}(2,v^{{\perp}})). \]

Since $i_v^{*}(\mathcal {U}^{\vee })$ is the dual tautological bundle and $i_v^{*}(\mathcal {S}_+)$ is the spinor bundle on the odd orthogonal Grassmannian $\operatorname {OG}(2,v^{\perp }) \cong \operatorname {OG}(2,2n-1)$, it follows from [Reference KuznetsovKuz08, Theorem 7.1] that there is a semiorthogonal decomposition

(3.36)\begin{equation} {{\mathbf{D}^{\mathrm{b}}}}(\operatorname{OG}(2,2n-1)) = \langle \mathcal{C} , \mathcal{C}(1) ,\ldots , \mathcal{C}(2n-5) \rangle. \end{equation}

Twisting it by $\mathcal {O}(2 - n)$, we obtain a semiorthogonal decomposition

(3.37)\begin{equation} {{\mathbf{D}^{\mathrm{b}}}}(\operatorname{OG}(2,2n-1)) = \langle \mathcal{C}(2-n),\ldots, \mathcal{C}({-}1), \mathcal{C} , \mathcal{C}(1) ,\ldots , \mathcal{C}(n-3) \rangle. \end{equation}

Thus, the definition of the set $\mathbf {E}'$ implies that the objects $i_v^{*}E$ for $E \in \mathbf {E}'$ generate the category ${{\mathbf {D}^{\mathrm {b}}}}(\operatorname {OG}(2,v^{\perp }))$, hence (3.35) implies that $i_v^{*}F = 0$. Since this holds for any non-isotropic $v$, it follows from Lemma 3.20 that $F = 0$.

Proof. To prove the claim it is enough to show the following two facts:

• – the object $F_i (i-1) [n-1+i]$ is contained in the orthogonal $\langle \mathcal {A}, \mathcal {A}(1),\ldots , \mathcal {A}(i) \rangle ^{\perp }$;

• – there exists a morphism $S^{n-2} \mathcal {U}^{\vee }(i) \to F_i(i-1)[n-1+i]$, whose cone lies in the category $\langle \mathcal {A}, \mathcal {A}(1),\ldots , \mathcal {A}(i) \rangle$.

Twisting (3.38) by $\mathcal {O}(i-1)$ and using the semiorthogonality of (3.9), we deduce the first of these facts. On the other hand, considering (3.39) twisted by $\mathcal {O}(i-1)$ as a Yoneda extension of the vector bundle $S^{n-2} \mathcal {U}^{\vee }(i)$ by $F_i(i-1)$ of length $n-1+i$, that is, as a morphism $S^{n-2} \mathcal {U}^{\vee }(i) \to F_i(i-1)[n-1+i]$, we conclude that its cone is quasi-isomorphic to the subcomplex of middle terms, hence belongs to the subcategory $\langle \mathcal {A}, \mathcal {A}(1),\ldots , \mathcal {A}(i) \rangle$. This proves the second fact.

Proof. We start with the first claim. Since we already know that (3.40) is an exceptional collection, we may assume $i < j$. From (3.39) twisted by $\mathcal {O}(i-1)$ we deduce

\[ F_i(i-1) \in \langle \mathcal{A} , \mathcal{A}(1) ,\ldots , \mathcal{A}(i-1), \mathcal{B}(i) \rangle, \]

and from (3.38) twisted by $\mathcal {O}(j-1)$ and with $i$ replaced by $j$ we deduce

(3.41)\begin{equation} F_j(j-1) \in \langle \mathcal{B}(j-n+1) , \mathcal{A}(j-n+2) ,\ldots , \mathcal{A}({-}1) \rangle. \end{equation}

Using an appropriate twist of (3.9), one easily sees that non-trivial $\operatorname {Ext}$s can only come from

\[ \operatorname{Ext}^{{\bullet}}(S^{n-2} \mathcal{U}^{{\vee}}(i), S^{n-2} \mathcal{U}^{{\vee}}(j-n+1)). \]

By Lemma 3.11 this space is non-zero (and is isomorphic to $\Bbbk [4-2n]$) only if $j = i + 1$ and the claim follows.

For the second claim, using the inclusion (3.41), we see that non-trivial $\operatorname {Ext}$s can only come from

\[ \operatorname{Ext}^{{\bullet}}(\mathcal{U}^{2\omega_{\varepsilon}}({-}1) , S^{n-2} \mathcal{U}^{{\vee}}(i-n+1)), \]

where $\varepsilon \in \{n-1, n\}$. By Lemma 3.14 and Serre duality this space is non-zero (and is isomorphic to $\Bbbk [2-n]$) only if $i = 1$, and the claim follows.