I am currently learning about algebraic viewpoint on closed embedded subsupermanifolds. In particular, I am struggling with something which should be ''easy to see''. Namely, I refer to the lemma just above Proposition 3.2.9 in "Introduction to the Theory of Supermanifolds" by D. A. Leites (http://iopscience.iop.org/0036-0279/35/1/R01).

Remark: I know there is an answer below. It, however, does not work for supermanifolds (I am not able to generalize).

The setting: Suppose we have a map $\psi: L \rightarrow M$ of supermanifolds, such that it is transversal to a given closed embedded subsupermanifold $N \subseteq M$.

Now, to every closed embedded subsupermanifold, one may assign a unique ideal $\mathcal{J}_{N} \leq C^{\infty}(M)$ in the superalgebra of its global functions, defined by $\mathcal{J}_{N} = \{ f \in C^{\infty}(M) \; | \; j^{*}(f) = 0 \}$, where $j: N \rightarrow M$ is the embedding. It can be then shown that $j^{\ast}: C^{\infty}(M) \rightarrow C^{\infty}(N)$ is a superalgebra epimorphism and $\mathcal{J}_{N}$ is its kernel, whence $C^{\infty}(N) \cong C^{\infty}(M) / \mathcal{J}_{N}$

Next, by definition of the map of supermanifolds, we have a superalgebra morphism $\psi^{\ast}: C^{\infty}(M) \rightarrow C^{\infty}(L)$. We can thus consider a subset $\psi^{\ast}(\mathcal{J}_{N}) \subseteq C^{\infty}(L)$. As $\psi^{\ast}$ is usually not surjective, this is in general not an ideal. We can, however, consider the ideal $\mathcal{I} = \langle \psi^{\ast}(\mathcal{J}_{N}) \rangle \leq C^{\infty}(L)$ generated by this subset.

THE ACTUAL QUESTION: $\mathcal{I}$ is supposed to have the following property: Let $\{ f_{\mu} \}_{\mu \in J}$ be any collection of functions in $\mathcal{I}$, such that $\{ supp(f_{\mu}) \}_{\mu \in J}$ is locally finite. Then also their sum $\sum_{\mu \in J} f_{\mu}$ must be function in $\mathcal{I}$.

I am stuck at this very point. According to D.A. Leites, this should be easy to see.

For the sake of completeness, let me recall some definitions:

A collection $\{C_{\mu} \}_{\mu \in J}$ of subsets of any topological space is locally finite, if for every compact subset $K$, $C_{\mu} \cap K \neq \emptyset$ only for finitely many $\mu \in J$. On (ordinary) manifolds, this is equivalent to every point $m$ having a neighborhood $U_{m}$, such that $C_{\mu} \cap U_{m} \neq \emptyset$ only for finitely many $\mu \in J$.

For a function $f$ on a supermanifold $M$, its support $supp(f)$ is a set of points $m$ of the underlying manifold $|M|$, where the germ $[f]_{m}$ of $f$ does not vanish.

Some comments: (i) The ideal $\mathcal{J}_{N}$ has this very property. For each $U \subseteq M$ open, let $j^{\ast}_{U}: C^{\infty}(U) \rightarrow C^{\infty}(U \cap N)$ be the superalgebra morphism induced by the pullback by $j$. For each point $m \in M$, we can pick a precompact neighborhood $U_{m}$. If $f_{\mu} \in \mathcal{J}_{N}$ for every $\mu \in J$, we obtain

$(j^{\ast}( \sum_{\mu \in J} f_{\mu} ))|_{U_{m} \cap N} = j^{\ast}_{U_{m}}( (\sum_{\mu \in J} f_{\mu} )|_{U_{m}}) = \sum_{\mu \in J} j^{\ast}_{U_{m}}( f_{\mu}|_{U_{m}}) = \sum_{\mu \in J} (j^{\ast}(f_{\mu}))|_{U_{m} \cap N} = 0,$

where it was important that after restriction to $U_{m}$, the sum is finite. But $m$ was arbitrary and $\{ U_{m} \cap N \}_{m \in M}$ forms an open cover of $N$, which proves that $j^{\ast}( \sum_{\mu \in J} f_{\mu} ) = 0$, that is $\sum_{\mu \in J} f_{\mu} \in \mathcal{J}_{N}$.

(ii) I don't know whether the transversality of $\psi$ to $N$ is somewhat important at this point.

(iii) D. A. Leites only assumes that the indexing set $J$ is countable. This is not very important though, as for general $J$, one can always find a countable subset $J'$, such that $\sum_{\mu \in J} f_{\mu} = \sum_{\mu' \in J'} f_{\mu'}$.


Update: I added a previous version (Version 2) which does not use dimension theory and tubular neighborhoods and it might be possible to generalize it to supermanifolds. However, it requires that $N$ can be covered by finitely many coordinate balls.

VERSION 1: FOR SMOOTH MANIFOLDS AND ANY N AND $\psi$ (uses that $N$ can be covered by $\mathrm{dim}(N)+1$ charts and that there is a tubular neighborhood of $N$)

It holds $$ \mathcal{I} = \Bigl\{ \sum_{i=1}^k (g_i \circ \psi) h_i \ \Bigl|\ k\in\mathbb{N}, g_i\in C^\infty(M), h_i\in C^\infty(L): g_i(N) = 0 \Bigr\}. $$ Given a locally finite collection of smooth functions $(f_i\in \mathcal{I} \mid i\in I)$, we will show that $$ \sum_{i\in I} f_i \in \mathcal{I}. $$

Lemma 1: We can assume that $f_i = (g_i \circ \psi) h_i$ for all $i\in\mathcal{I}$.

Proof: Pick a partition of unity $(\eta_j \mid j\in \mathcal{J})$ on $L$ such that $\eta_j$ has compact support for every $j\in\mathcal{J}$. Let $\mathcal{A}:= \mathcal{I}\times\mathcal{J}$, and define $$ f_\alpha:= \eta_j f_i $$ for all $\alpha=(i,j)\in \mathcal{A}$. The system $(f_{\alpha}\mid \alpha\in\mathcal{A})$ is locally finite, and it holds $$ \sum_{\alpha\in\mathcal{A}} f_{\alpha} = \sum_{j\in \mathcal{J}} \eta_j \sum_{i\in\mathcal{I}} f_i = \sum_{i\in\mathcal{I}} f_i. $$ For every $i\in\mathcal{I}$, there is an $m_i\in \mathbb{N}$ such that $f_i = \sum_{l=1}^{m_i} (g_{il}\circ\psi)h_{il}$. It follows that for every $\alpha = (i,j)\in\mathcal{A}$, it holds $$ f_\alpha = \eta_j f_i = \sum_{l=1}^{m_i} (g_{il}\circ\psi)(\eta_j h_{il}) = \sum_{i=1}^{m_i} (g_{il}\circ\psi)h_{\alpha l} \in \mathcal{I}, $$ where we defined $$ h_{\alpha l} := \begin{cases} 0 & \text{if }f_\alpha=0, \\ \eta_j h_{il} & \text{otherwise.} \end{cases} $$ Let $x\in L$. There is an open neighborhood $U$ of $x$ and a finite subset $\mathcal{J}_0\subset \mathcal{J}$ such that $\mathrm{supp}(\eta_j) \cap U = 0$ for all $j\in \mathcal{J}\backslash\mathcal{J}_0$. Because $\sum_{j\in \mathcal{J}}\eta_j = 1$, it holds $U\subset \bigcup_{j\in \mathcal{J}_0} \{ x\in M \mid \eta_j(x)\neq 0\}$. Because $\bigcup_{j\in \mathcal{J}_0} \mathrm{supp}(\eta_j)$ is compact, there exists a finite subset $\mathcal{I}_0\subset \mathcal{I}$ such that $$ \mathrm{supp}(f_i)\cap \bigcup_{j\in \mathcal{J}_0} \mathrm{supp}(\eta_j) = \emptyset $$ for all $i\in \mathcal{I}\backslash\mathcal{I}_0$. Suppose that $\mathrm{supp}(h_{\alpha l})\cap U \neq 0$ for some $\alpha = (i,j)$ and $l\in \{1,\dotsc,m_i\}$. Because $\mathrm{supp}(h_{\alpha l})\subset\mathrm{supp}(\eta_j)$, it must hold $j\in \mathcal{J}_0$. By definition of $h_{\alpha l}$, it holds either $h_{\alpha l} = 0$ if $f_\alpha = 0$, which is equivalent to $\{x\in M \mid f_i(x)\neq 0\}\cap\{x\in M\mid \eta_j(x)\neq 0\}=\emptyset$ which is equivalent to $\mathrm{supp}(f_i)\cap\{x\in M\mid \eta_j(x)\neq 0\} = \emptyset$, or $h_{\alpha_l} = \eta_j h_{il}$. The second option can possibly occur only for $i\in\mathcal{I}_0$. This shows that the collection $$ ((g_{il}\circ\psi)h_{\alpha l} \mid \alpha=(i,j)\in\mathcal{A}, l\in\{1,\dotsc,m_i\}) $$ is locally finite. Its sum equals $\sum_{\alpha\in\mathcal{A}} f_\alpha$ and hence $\sum_{i\in \mathcal{I}}f_i$ by construction. QED

By Lemma 1, we can assume that $f_i = (g_i\circ\psi)h_i$ for $g_i\in C^\infty(M)$ with $g_i(N)=0$ and $h_i\in C^\infty(L)$ without lost of generality.

Denote $k:=\dim(N)$ and $n:=\dim(M)$. Pick a tubular neighborhood $\mathcal{N}(N)$ of $N$ in $M$. The $k$-dimensional manifold $N$ can be always covered by $k+1$ (not necessarily connected) charts $U_1$, $\dotsc$, $U_{k+1}$. Every chart $U_j$ on $N$ induces a submanifold chart $V_j = \mathcal{N}(U_j)$ on $M$. Let $V_0\subset M$ be an open subset disjoint from $N$ such that $M = \cup_{j=0}^{k+1} V_j$. Let $\lambda_0$, $\dotsc$, $\lambda_{k+1}$ be a subordinate partition of unity. Let $\mu$ be a bump function which equals $1$ on $\mathrm{supp}(\lambda_0)$ and vanishes on $N$.

Let $(x_j,y_j)\in \mathbb{R}^n$ be coordinates on $\mathcal{N}(U_j)$ such that $x_j = (x_j^1,\dotsc,x_j^k)$ gives coordinates on the base and $y_j = (y_j^1,\dotsc,y_j^{n-k})$ on fibers. An important feature of $\mathcal{N}(U_j)$ is that it contains the vertical line $\gamma(t) = (x_j,0) + t((x_j,y_j)-(x_j,0))$ connecting $(x_j,0)$ and $(x_j,y_j)$. The Fundamental theorem of calculus in the form $$ f(\gamma(1))-f(\gamma(0)) = \int_{0}^1 (\nabla f)(\gamma(t))\cdot\gamma'(t) \mathrm{d}t $$ then asserts that the following holds for all $i\in I$ and $j\in \{1,\dotsc,k+1\}$ on the entire $\mathcal{N}(U_j)$: $$ (\lambda_j g_i)(x_j,y_j) - \underbrace{(\lambda_j g_i)(x_j,0)}_{=0} = \sum_{a=1}^{n-k} y^a_j \underbrace{\int_{0}^1 \frac{\partial(\lambda_j g_i)}{\partial y^a_j}(x_j,ty_j) \mathrm{d}t}_{\displaystyle=:u_{i a}^j}. $$ Let $\tilde{y}^a_j$ and $\tilde{u}_{ia}^j$ be the smooth functions on $M$ obtained from $y^a_j$ and $u_{ia}^j$, respectively, by multiplication with a bump function which is $1$ on $\mathrm{supp} \lambda_j$ and $0$ on a neighborhood of the closure of the complement of $\mathcal{N}(U_j)$.

For all $i\in I$ and $j\in \{1,\dotsc,k+1\}$, we have the following relations on $M$: $$ \lambda_0 g_i = \mu \lambda_0 g_i\quad\text{and}\quad\lambda_j g_i = \sum_{a=1}^{n-k} \tilde{y}^a_j \tilde{u}_{ia}^j. $$ Using this, we compute \begin{align*} \sum_{i\in I} (g_i \circ \psi) h_i &= \sum_{i\in I} \sum_{j=0}^{k+1} (\lambda_j g_i \circ \psi) h_i \\ & = \sum_{i\in I} (\lambda_0 g_i \circ \psi) h_i + \sum_{i\in I} \sum_{j=1}^{k+1} \sum_{a=1}^{n-k} (\tilde{y}^a_j \circ \psi)(\tilde{u}_{ia}^j \circ \psi)h_i \\ & = (\mu\circ\psi)\sum_{i\in I}(\lambda_0 g_i \circ \psi) h_i+ \sum_{j=1}^{k+1} \sum_{a=1}^{n-k} (\tilde{y}^a_j\circ \psi) \sum_{i\in I} (\tilde{u}_{ia}^j \circ \psi)h_i\\ & = (G_0 \circ \psi) H_0 + \sum_{j=1}^{k+1} \sum_{a=1}^{n-k} (G_{ja}\circ\psi)H_{ja}, \end{align*} where we denoted $$ G_0:= \mu,\quad G_{ja}:=\tilde{y}^a_j,\quad H_0:=\sum_{i\in I}(\lambda_0 g_i \circ \psi) h_i,\quad H_{ja}:=\sum_{i\in I} (\tilde{u}_{ia}^j\circ \psi)h_i. $$ It holds $G_0$, $G_{ja}\in C^\infty(M)$, $G_0(N)=G_{ja}(N) = 0$, $H_0$, $H_{ja}\in C^\infty(L)$, and it follows that $\sum_{i\in I} (g_i \circ \psi) h_i \in \mathcal{I}$.

VERSION 2: PROOF WHEN $N$ CAN BE COVERED BY FINITELY MANY COMPATIBLE COORDINATE BALLS (not using dimension theory and tubular neighborhood)

Write $\mathbb{R}^n = \mathbb{R}^k\times \mathbb{R}^{n-k}$ with coordinates $(x,y)$. Let $f: \mathbb{R}^n\rightarrow \mathbb{R}$ be a smooth function vanishing at $\{(x,y) \mid x = 0\}$. Then the fundamental theorem of calculus asserts that the following holds for all $(x,y)\in \mathbb{R}^n$: $$ f(x,y) = \sum_{j=1}^{k} x^j \int_{0}^1 \frac{\partial f}{\partial x^j}(tx,y) dt. $$ Let $U_\alpha$ $(\alpha\in\mathcal{A})$ be a cover of $N$ by coordinate balls, and let $\lambda_\alpha$ $(\alpha\in\mathcal{A})$ be a subordinate partition of unity. Suppose that we are given $\sum_{i\in I} (g_i \circ \psi) h_i$ as above and we want to show that it lies in $\mathcal{I}$. We can even assume that the support lies in an arbitrary small neighborhood of $\psi^{-1}(N)$. Using the analytical fact above, there are smooth functions $x_{\alpha}^j$ vanishing on $N$ and smooth functions $u^{\alpha}_{ij}$ for all $i\in I$, $\alpha\in\mathcal{A}$ and $j\in\{1,\dotsc,k:=\mathrm{codim} N\}$ such that $$ \lambda_\alpha g_i = \sum_{j=1}^k x_\alpha^j u_{ij}^\alpha. $$ We compute \begin{align*} \sum_{i\in I} (g_i \circ \psi) h_i = \sum_{j=1}^k \sum_{\alpha\in\mathcal{A}} (\lambda_\alpha x^j_\alpha\circ\psi) \sum_{i\in I} (u_{ij}^\alpha \circ \psi) h_i. \end{align*} If $\mathcal{A}$ is finite, then we are done.

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  • $\begingroup$ Your answer related to the original formulation of this question (I have edited it significantly). In a summary - you give an answer for the case where all the manifolds are ordinary manifolds. However, in supermanifolds, one does not have a full apparautus of tubular neighborhoods, or the fact that k-dimensional manifold can be covered by k+1 charts. $\endgroup$ – Jan Vysoky Apr 21 at 13:53
  • $\begingroup$ Version 2 can be significantly simplified. By the modification of Lemma 1 above, you can assume that $f_{i} = (g_{i} \circ \psi) h_{i}$, where $\{ supp(g_{i}) \}_{i \in I}$ is locally finite. It then suffices to cover $N$ by finitely many precompact sets $\{U_{\alpha} \}_{\alpha=1}^{m}$ (e.g. coordinate balls). Let $U_{0} := M - N$ and let $\{ \lambda_{\alpha} \}_{\alpha = 0}^{m}$ be the corresponding partition of unity. Note that $\lambda_{0} \in \mathcal{J}_{N}$. Now, as $U_{\alpha}$ are compact for $\alpha > 0$, there is a finite subset $I_{\alpha} \subseteq I$ such that $\endgroup$ – Jan Vysoky Apr 28 at 12:52
  • $\begingroup$ such that $\lambda_{\alpha} \cdot g_{i} \neq 0$ only for $i \in I_{\alpha}$. Also note that $\lambda_{\alpha} \cdot g_{i} \in \mathcal{J}_{N}$. Then $\sum_{i \in I} (g_{i} \circ \psi) h_{i} = \sum_{i \in I} ((\sum_{\alpha=0}^{m}\lambda_{\alpha}) \cdot g_{i} \circ \psi) h_{i} = (\sum_{i \in I} (g_{i} \circ \psi) h_{i}) \cdot (\lambda_{0} \circ \psi)$ $+ \sum_{\alpha=1}^{m} \sum_{i \in I_{\alpha}} (\lambda_{\alpha} \cdot g_{i} \circ \psi) h_{i}$. The first summand is in $\mathcal{I}$ as $\lambda_{0} \in \mathcal{J}_{N}$ and the rest is a finite sum of elements in $\mathcal{I}$. $\endgroup$ – Jan Vysoky Apr 28 at 13:02

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