The followings are from Mnev's paper about BV formalism.

Example 4.15 (Definition of split supermanifold)

Let $E \to M$ be a rank $m$ vector bundle over $n$-manifold $M$, then there exists a split $(n|m)$-supermanifold $\Pi E$ with body $M$ and structure sheaf $\mathcal O_{\Pi E} = \Gamma(M ,\bigwedge^{\bullet}E^*)$.

Section 4.2.4 (Definition of Berezin line bundle of split supermanifold)

Let $\mathcal M = \Pi E$ be the split $(n|m)$-supermanifold of vector bundle $E \to M$.
Berezin line bundle of $\mathcal M$ is $\text{Ber}(\mathcal M) := \bigwedge^{n} T^*M\otimes \bigwedge^m E$ over $M$.

For $T^*M \to M$, we have $\text{Ber}(\Pi T^*M) \cong (\bigwedge^{n} T^*M)^{\otimes 2}$. This is (161).

Then in (162), it writes

Similarly, for $\mathcal N$ a supermanifold, one has $\text{Ber}(\Pi T^*\mathcal N)|_{\mathcal N} \cong\text{Ber}(\mathcal N)^{\otimes 2}$.
Here we understand $\text{Ber}(\mathcal N)$ as a line bundle over $\mathcal N$ and the l.h.s. is a pullback of a line bundle over $\Pi T^* \mathcal N$ to $\mathcal N$.

I have difficulty in understanding the above sentence.

Suppose $\mathcal N$ is a $(k|n-k)$-supermanifold of body $N$, then $T^*\mathcal N$ has dimension $(2k|n-2k)$ and $\Pi T^*\mathcal N$ has dimension $(n|n)$.

Question 1:

What is the expression of $\text{Ber}(\Pi T^*\mathcal N)$ for supermanifold $\mathcal N$?

Question 2:

From Batchelor's theorem, we can assume $\mathcal N = \Pi{B}$ for some bundle $B \to N$.

By definition, $\text{Ber}(\mathcal N) = \bigwedge^{k}T^*N \otimes \bigwedge^{n-k} B$, so how can we show $$\text{Ber}(\Pi T^*\mathcal N)|_{\mathcal N} \cong \text{Ber}(\mathcal N)^{\otimes 2} ?$$

Update: My senior told me that this is a conclusion of adjunction formula for supermanifold, though I'm not very familiar with this and I can't understand the followings. Here's the proof:

$\text{Ber}$ can be defined on any locally free sheaf. Especially, when we say $\text{Ber}(X)$ we actually mean $\text{Ber}(T^*_X)$, here $T^*_X$ means cotangent sheaf on $X$.

$X \hookrightarrow Y$ is closed embedding of smooth supermanifold, ideal of $X$ is $\mathcal I$, then $$0 \to \mathcal I/\mathcal I^2 \to \Omega^{1}_{Y}|_{X} \to \Omega^{1}_{X} \to 0$$ is exact. This implies $\text{Ber}(Y)|_{X} \cong \text{Ber}(X) \otimes \text{Ber}(\mathcal I / \mathcal I^2)$.

Take $Y = V(\Pi T_{X})$ (here $V(E)$ means $E^*$ bundle and $T_X$ means tangent sheaf), then $\mathcal I/\mathcal I^2 = \Pi T_X$ and $$\text{Ber}(\Pi T_X) = (\text{Ber}(T_X))^{-1} = \text{Ber}(X)$$ here $(\text{Ber}(T_X))^{-1}$ means dual bundle of $\text{Ber}(T_X)$.

Or is there any reference for this?

I really can't find too much information of Berezin bundle.

Thanks for your time and effort.


The Berezinian of a vector bundle $E$ arises via an associated bundle construction: it is reasonably easy to check that $$ \begin{pmatrix}A&B\\C&D\end{pmatrix}\mapsto |\operatorname{det}(A)|\operatorname{det} ^{-1}(D- CA^{-1}B) $$ defines a super Lie group homomorphism $\operatorname{Ber}:GL(\mathbb R^{m|n})\to\mathbb R^\times$. (Here $A\in GL(\mathbb R^n),B\in\mathbb R^{m\times n},C\in\mathbb R^{n\times m},D\in GL(\mathbb R^n)$ are the even-to-even etc. blocks of an element of a matrix in $GL(\mathbb R^{m|n})$). It arises by integration of the supertrace, which is a super Lie algebra homomorphism $\mathfrak{gl}(\mathbb R^{m|n})\to \mathbb R$. Then the Berezinian of $E$, thought of as a locally free sheaf over the base supermanifold, can be defined as the sheafification of the presheaf which assigns to an open subsupermanifold $U$ the $\operatorname{sMan}(U,GL(\mathbb R^{m|n}))$-equivariant functions from trivializations of $E|_U$ to $\mathcal O(U)$. In words, a trivialization of $E$ induces a trivialization of $\operatorname{Ber}(E)$, and a different trivialization of $E$ changes this trivialization by the Berezinian of the matrix-valued "gauge transformation" between them.

Your advisor's statement is that a short exact sequence of vector bundles $0\to F_1\to E\to F_2$ gives rise to a natural isomorphism of line bundles $\operatorname{Ber}(E)\cong \operatorname{Ber}(F_1)\otimes \operatorname{Ber}(F_2)$. In terms of the representation, such a short exact sequence is the same as a reduction of structure group from $GL(\mathbb R^{m|n})$ to the group of upper triangular matrices $\big(GL(\mathbb R^{m_1|n_1})\times GL(\mathbb R^{m_2|n_2})\big)\ltimes \mathbb R^{m_1|n_1\times m_2|n_2}$. Restricted to this subgroup, the Berezinian equals the product of the Berezinians of the two blocks, which under the associated bundle construction corresponds to this isomorphism. There are some fermion sign issues when switching the $F_i$; these become the standard Koszul signs if we stipulate that the parity of the Berezinian super line equals that of the odd dimension of the superspace.

Given any vector bundle $E\to X$ over a super manifold, we can form the total space which is also donated by $E$ and comes with a canonical map $p:E\to X$. Then there is a short exact sequence $$ 0\to p^* E\to TE\xrightarrow{p_*} p^* TX $$ which gives rise to an isomorphism $\operatorname{Ber}(TE)\cong p^*\big(\operatorname{Ber}(E)\otimes \operatorname{Ber}(TX)\big)$; note that this isomorphism is only invariant under diffeomorphisms of the total space $E$ which preserve this exact sequence. This gives the formula for the Berezinian of a split supermanifold which you give in your question 2. However, when we pull back this exact sequence to $X$ along the zero section $s$, it is canonically split. For your question about the shifted cotangent bundle, this gives $\operatorname{Ber}(TX)\otimes \operatorname{Ber}(\Pi T^*X)$. Now it is straightforward to check at the level of representations that both changing the parity and dualizing replace the Berezinian by its dual, so that $\operatorname{Ber}(\Pi T^*X)\cong \operatorname{Ber}(TX)$, which then gives the isomorphism in (162). In general, since Batchelor's theorem does not give a natural split form for a supermanifold, it is usually not straightforward to give a natural construction using it, so I don't have an answer to your question 2.

Finally, there is a good reason why the Berezinian of the shifted cotangent bundle is the square of a line bundle: it carries a canonical odd symplectic structure, i.e. a reduction of structure group from $GL(\mathbb R^{n|n})$ to the group

$$ Sp(\mathbb R^{n|n}) = \left\{\begin{pmatrix}A&B\\ C&D\end{pmatrix}\mid A^TC=C^TA, D^TB=-B^TD, A^TD - C^TB = I\right\} $$ It's straightforward to show that $(A^{-1})^T = D - CA^{-1}B$, so that $\operatorname{Ber}\begin{pmatrix}A&B\\ C&D\end{pmatrix}=\operatorname{det}^2(A)$ has a canonical square root. Under the associated bundle construction, this gives rise to the half-Berezinian line bundle, whose restriction to any Lagrangian submanifold is canonically identified with the Berezinian of that submanifold. A reference for the last part is Section 4.6 of Pavel Mnev's book "Quantum Field Theory: Batalin–Vilkovisky Formalism and Its Applications".


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