2 Setup and background

In this section, we recall the main concepts and results for the rest of the paper. For the basic definitions, see [Reference Ishii12, Reference de Jong and Pfister13].

Definition 2.1 Let $\Omega \subset \mathbb {C}^n$ be an open subset. A closed subset $X \subset \Omega $ is called an analytic subset (or analytic set) of $\Omega $ if for all $x\in X$ , there is an open neighborhood $V\subset \Omega $ of x and a finite set of analytic functions $f_1,\dots ,f_s\in \mathcal {O}_n(\Omega )$ defined on V such that

$$ \begin{align*}X\cap V =\{x\in V \ \mid \ f_1(x)=\dots=f_s(x)=0\}.\end{align*} $$

A pair $(\varphi , \varphi ^{\ast }): (X,\mathcal {O}_X)\to (Y,\mathcal {O}_Y)$ is called a morphism of ringed spaces if the map $\varphi : X\to Y$ is continuous and $\varphi ^*: \mathcal {O}_Y\to \varphi _*\mathcal {O}_X$ is a morphism of sheaves of rings. Also, $\varphi _*\mathcal {O}_X$ is the sheaf of commutative rings given by $\varphi _*\mathcal {O}_X(U)=\Gamma (\varphi ^{-1}(U),\mathcal {O}_X)$ , for any open subset $U\subset Y$ .

A morphism of locally ringed spaces $(X, \mathcal {O}_X)$ and $(Y,\mathcal {O}_Y)$ is a morphism of ringed spaces $(\varphi , \varphi ^{\ast }): (X, \mathcal {O}_X)\to (Y,\mathcal {O}_Y)$ such that for all $x\in X$ , the homomorphism $\varphi ^*_x: \mathcal {O}_{Y,\varphi (x)}\to \mathcal {O}_{X,x}$ induced from $\varphi ^*$ is a local homomorphism, i.e., ${\varphi ^*}^{-1}(\mathfrak {m}_{X,x})=\mathfrak {m}_{Y,\varphi (x)}$ .

A morphism $(\varphi , \varphi ^{\ast }): (X, \mathcal {O}_X)\to (Y,\mathcal {O}_Y)$ is an isomorphism if $\varphi $ is a homeomorphism and $\varphi ^*$ is an isomorphism of sheaves of rings.

From now on, the ringed space $(X,\mathcal {O}_X)$ and the morphism of ringed spaces $(\varphi ,\varphi ^{\ast })$ will be denoted by X and $\varphi $ , respectively.

Remark 2.6 It should be noted that the elements of the stalks $\mathcal {O}_{X,x}$ are seen as germs at x of holomorphic functions on X. Each germ is represented by a holomorphic function $f\in \mathcal {O}_X(U)$ , defined on an open neighborhood U of x. Conversely, each $f\in \mathcal {O}_X(U)$ defines a unique germ at $x\in U$ , which is denoted by $f_x$ . Hence, since $(X,\mathcal {O}_X)$ is an analytic space and $x=(a_1,\dots ,a_n)\in X\subset \Omega \subset \mathbb {C}^n$ , one has the isomorphism

$$ \begin{align*}\mathcal{O}_{X,x}\cong \mathcal{O}_{\mathbb{C}^n,x}/\mathcal{I}_{X,x}\cong \mathbb{C}\{x_1-a_1,\dots, x_n-a_n\}/\mathcal{I}_{X,x},\end{align*} $$

where $\mathcal {I}_{X,x}=\{f_x\in \mathcal {O}_{\mathbb {C}^n,x}\,\,|\,\,\exists \ f\in \mathcal {O}_{\mathbb {C}^n}(U)\,\, \mathrm {representing }\,\, f_x \,\, \mathrm {and }\,\, f|_{U\cap X}=0 \}.$ Now, the fact that $\mathcal {O}_{\mathbb {C}^n,x}$ is Noetherian gives that the ideal $\mathcal {I}_{X,x}$ is finitely generated, and so there exists $f_1,\dots , f_k\in \mathcal {O}_{\mathbb {C}^n,x}$ such that $\mathcal {I}_{X,x}=\langle f_1,\dots , f_k \rangle $ . For this paper, $\mathcal {I}_{X,x}$ is an ideal that defines the germ $(X,x)$ of an analytic space. Note that $\mathcal {O}_{X,x}$ is an analytic $\mathbb {C}$ -algebra and is a local ring with maximal ideal $\mathfrak {m}_{X,x}=\{f\in \mathcal {O}_{X,x} \ | \ f(x)=0\}$ .

Set $X\coprod Y$ as the co-product or disjoint union of sets X and Y.

Definition 2.7 Let $\alpha : Z \to X$ and $\beta : Z \to Y$ be morphisms of ringed spaces. Set

$$ \begin{align*}X \sqcup_Z Y= X\coprod Y/\sim,\end{align*} $$

where the relation $\sim $ is generated by relations of the form $x\sim y$ ( $x\in X$ , $y\in Y$ ), provided there exists $z\in Z$ such that $\alpha (z) = x$ and $\beta (z)=y$ .

Namely, it is the smallest equivalence relation on $X\coprod Y$ such that after passing to the quotient $X\coprod Y/\sim $ the following square becomes commutative

(2.1)

where f and g are the continuous natural maps.

Since $(X,\mathcal {O}_X)$ , $(Y,\mathcal {O}_Y)$ , and $(Z,\mathcal {O}_Z)$ are ringed spaces, [Reference Schwede20, Proposition 2.2] provides that $(X\sqcup _ZY, \mathcal {O}_{X\sqcup _ZY})$ is also a ringed space, and so f and g becomes morphisms of ringed spaces. Also, note that this definition satisfies the universal property by [Reference Schwede20, Theorem 2.3].

Analogous to the morphisms of germs $\alpha _z: (Z,z) \to (X,\alpha (z))$ and $\beta _z: (Z,z) \to (Y,\beta (z))$ , the previous definition can be made for germs of analytic spaces $(X,x)$ , $(Y,y)$ , and $(Z,z)$ , and denoted by $(X,\alpha (z))\sqcup _{(Z,z)}(Y,\beta (z))$ . In the rest of the paper $(X,\alpha (z))\sqcup _{(Z,z)}(Y,\beta (z))$ will be denoted by $(X,x)\sqcup _{(Z,z)}(Y,y)$ , where $\alpha (z)=x$ and $\beta (z)=y$ . When a germ $(Z,z)$ is a reduced point, i.e., $(Z,z) =(z,z)$ , we will denote $(Z,z)$ by $\{z\}$ .

Now, we recall an important definition for this paper.

Definition 2.8 The fiber product of homomorphisms $\alpha _z^{\ast }:\mathcal {O}_{X,x} \to \mathcal {O}_{Z,z}$ , $\beta _z^{\ast }:\mathcal {O}_{Y,y} \to \mathcal {O}_{Z,z}$ of $\mathbb {C}$ -algebras is defined by

$$ \begin{align*}\mathcal{O}_{X,x} \times_{\mathcal{O}_{Z,z}} \mathcal{O}_{Y,y}:=\{(s,t)\in \mathcal{O}_{X,x} \times \mathcal{O}_{Y,y} \ \mid \ \alpha_z^{\ast}(s)=\beta_z^{\ast}(t) \}.\end{align*} $$

By [Reference Ananthnarayan, Avramov and Moore2, Lemma 1.2], the fiber product is also a commutative and local ring with maximal ideal given by $\mathfrak {m} = \mathfrak {m}_{X,x} \times _{\mathfrak {m}_{Z,z}} \mathfrak {m}_{Y,y}$ , where $\mathfrak {m}_{X,x}$ , $\mathfrak {m}_{Y,y}$ and $\mathfrak {m}_{Z,z}$ are the maximal ideals of $\mathcal {O}_{X,x}$ , $\mathcal {O}_{Y,y}$ , and $\mathcal {O}_{Z,z}$ , respectively. Also it is a subring of $\mathcal {O}_{X,x} \times \mathcal {O}_{Y,y}$ and universal with respect to the commutative diagram

(2.2)

where $\pi _1(s,t) = s$ and $\pi _2(s,t)=t$ are natural surjections. Also in [Reference Ananthnarayan, Avramov and Moore2, Section 1(1.0.3)] and [Reference Endo, Goto and Isobe5, Lemma 2.1] the authors have shown that $\mathcal {O}_{X,x} \times _{\mathcal {O}_{Z,z}} \mathcal {O}_{Y,y}$ is a Noetherian local ring if both $\alpha _z^{\ast }$ and $\beta _z^{\ast }$ are surjective maps. It is important to realize that the assumptions over the maps are crucial for the Noetherianess of the fiber product ring [Reference Freitas, Jorge Perez and Miranda6, Example 2.9]. Also, if $(X,x)$ , $(Y,y)$ and $(Z,z)$ are germs of analytic spaces, then $\mathcal {O}_{X,x} \times _{\mathcal {O}_{Z,z}} \mathcal {O}_{Y,y}$ is a reduced ring [Reference Celikbas3, Proposition 4.2.18]. For the fiber product $\mathcal {O}_{X,x} \times _{\mathcal {O}_{Z,z}} \mathcal {O}_{Y,y}$ , we assume that $\mathcal {O}_{X,x}\neq \mathcal {O}_{Z,z}\neq \mathcal {O}_{Y,y}$ . Note that every $\mathcal {O}_{X,x}$ -module (or $\mathcal {O}_{Y,y}$ -module) is an $\mathcal {O}_{X,x} \times _{\mathcal {O}_{Z,z}} \mathcal {O}_{Y,y}$ -module via Diagram 2.2.

Remark 2.9 [Reference Freitas, Jorge Perez and Miranda6, Remark 2.11]

Let $(X,x)\subset (\mathbb {C}^n,x)$ and $(Y,y)\subset (\mathbb {C}^m,y)$ be two germs of analytic spaces, where $x=(a_1,\ldots ,a_n)$ and $y=(b_1,\ldots ,b_m)$ . Let $\mathcal {I}_{X,x}$ and $\mathcal {I}_{Y,y}$ be defining ideals of $(X,x)$ and $(Y,y)$ , respectively. Consider $R=\frac {\mathcal {O}_{n+m, (x,y)}}{(\mathcal {I}_{X,x}+\mathcal {I}_{Y,y}+((x_i-a_i)(y_j-b_j)))}$ , $i=1,\ldots , n$ and $j=1,\ldots , m$ , and let $I=(x_1-a_1,\ldots ,x_n-a_n)$ and $J=(y_1-b_1,\ldots ,y_m-b_m)$ be two ideals of R. Note that $I\cap J=0$ and therefore

$$ \begin{align*}\frac{\mathcal{O}_{n+m, (x,y)}}{(\mathcal{I}_{X,x}+\mathcal{I}_{Y,y}+((x_i-a_i)(y_j-b_j)))} \cong \mathcal{O}_{X,x}\times_{\mathbb{C}} \mathcal{O}_{Y,y}. \end{align*} $$

In particular, $\mathcal {O}_{X,x}\times _{\mathbb {C}} \mathcal {O}_{Y,y}$ is an analytic $\mathbb {C}$ -algebra and the ideal $\mathcal {I}_{X,x}+\mathcal {I}_{Y,y}+((x_i-a_i)(y_j-b_j))$ defines $(X,x)\sqcup _{\{z\}}(Y,y)$ .

Below we summarize the key results shown in [Reference Freitas, Jorge Perez and Miranda6] which establishes the good structure of the gluing of complex analytic space germs.

Lemma 2.10 [Reference Freitas, Jorge Perez and Miranda6, Proposition 2.10]

Let $\alpha : Z \to X$ and $\beta : Z \to Y$ be holomorphic mappings of analytic spaces. Then,

$$ \begin{align*}{\mathcal{O}}_{(X,\alpha(z))\sqcup_{(Z,z)}(Y,\beta(z))}\cong \mathcal{O}_{X,\alpha(z)} \times_{\mathcal{O}_{Z,z}} \mathcal{O}_{Y,\beta(z)}.\end{align*} $$

3 Poincaré series and Betti numbers of gluing of germs of analytic spaces

The main focus of this section is to define new classes of gluing of germs of complex analytic spaces and give the shape of their Poincaré series and Betti numbers. For this purpose, two important definitions are necessary.

Definition 3.1 Let $(X,x)\subset (\mathbb {C}^n,x)$ be a germ of an analytic space. Let M be a finitely generated $\mathcal {O}_{X,x}$ -module. The Poincaré series of M is given by

$$ \begin{align*}P_{M}^{\mathcal{O}_{X,x}}(t):= \sum_{i\geq 0}\dim_{\mathbb{C}} \mathrm{Tor}_i^{\mathcal{O}_{X,x}}\left(M, \mathbb{C} \right) t^i,\end{align*} $$

where $\mathbb {C}:= \frac {\mathcal {O}_{X,x}}{\mathfrak {m}_{X,x}}$ is the residue field. The number $\beta _i^{\mathcal {O}_{X,x}}(M):=\dim _{\mathbb {C}} \mathrm {Tor}_i^{\mathcal {O}_{ X,x}}\left (M, \mathbb {C}\right )$ is called i-th Betti number of M. Let I be an ideal of $\mathcal {O}_{X,x}.$ The Poincaré series of $\mathcal {O}_{X,x}/I$ is denoted by

$$ \begin{align*}P_{(Z,z)}^{(X,x)}(t):=P_{\mathcal{O}_{X,x}/I}^{\mathcal{O}_{X,x}}(t),\end{align*} $$

where $(Z,z)$ is subspace of $(X,x)$ defined by the reduced ideal I of $\mathcal {O}_{X,x}$ . The i-th Betti number of $(Z,z)$ is defined by $\beta _i^{(X,x)}(Z,z):=\beta _i^{\mathcal {O}_{X,x}}\left (\frac {\mathcal {O}_{X,x}}{I}\right ).$

Remark 3.2 Let $(Z,z)$ be a subspace of $(X,x)$ defined by the reduced ideal I. Set $\mu (Z,z)$ as the minimal number of generators of $\mathcal {O}_{X,x}/I$ . Then

$$ \begin{align*}P^{(X,x)}_{(Z,z)}(t)=\mu(Z,z)+tP^{(X,x)}_{(\Omega_1, \omega_1)}(t),\end{align*} $$

where $(\Omega _1, \omega _1)$ is the subspace that represents the first syzygy of $\mathcal {O}_{X,x}/I$ over $\mathcal {O}_{X,x}$ (see [Reference Dress and Krämer4]).

The next definition is motivated by the work of Levin [Reference Levin15].

Definition 3.3 Let $f: (Y,y) \to (X,x)$ be a morphism of germs of complex analytic spaces, such that the induced map $\mathcal {O}_{X,x}\to \mathcal {O}_{Y,y}$ is a surjective homomorphism. Then f is said to be large provided, for any $(Z,z)$ subspace of $(Y,y)$ considered as a subspace of $(X,x)$ , the following equality happens

$$ \begin{align*}P_{(Z,z)}^{(X,x)}=P_{(Z,z)}^{(Y,y)}P_{(Y,y)}^{(X,x)}.\end{align*} $$

Now, we are able to define new classes of gluing of germs of analytic spaces.

It easy to see that every strongly large gluing is large and therefore weakly large. The next example and remark show that these new classes of gluing of germs of analytic spaces are non-empty and contain interesting types of gluing.

We pose the following conjecture.

The next result is a key ingredient for the rest of the paper and shows the explicit shape of the Poincaré series of certain gluing of germs of analytic spaces.

Proof (i) The exact sequence

(3.1) $$ \begin{align} 0\longrightarrow \mathrm{ker}(\alpha_z^\ast)\longrightarrow \mathcal{O}_{X,x}\times_{\mathcal{O}_{Z,z}}\mathcal{O}_{Y,y}\overset{\pi_2}{\longrightarrow} \mathcal{O}_{Y,y}\longrightarrow 0 \end{align} $$

and Remark 3.2 gives

(3.2) $$ \begin{align} P_{(Y,y)}^{{(X,x)} \sqcup_{(Z,z)} {(Y,y)}}(t)=1+tP^{{(X,x)} \sqcup_{(Z,z)} {(Y,y)}}_{(K,k)}(t), \end{align} $$

where $(K,k)$ is the subspace of $(X,x)$ that represents the kernel of the map $\alpha _z^\ast $ . Since ${(X,x)\sqcup _{(Z,z)}(Y,y)}$ is weakly large, one obtains

(3.3) $$ \begin{align} tP^{{(X,x)} \sqcup_{(Z,z)} {(Y,y)} }_{(K,k)}(t) &= tP_{(K,k)}^{(X,x)}(t)P_{(X,x)}^{{(X,x)} \sqcup_{(Z,z)} {(Y,y)}}(t) \notag\\ &=\left(P_{(Z,z)}^{(X,x)}(t)-1\right)P^{{(X,x)} \sqcup_{(Z,z)} {(Y,y)}}_{(X,x)}(t), \end{align} $$

where the last equality follows by the exact sequence

(3.4) $$ \begin{align} 0\longrightarrow \mathrm{ker}(\alpha_z^\ast)\longrightarrow \mathcal{O}_{X,x}\overset{\mathrm{ker}(\alpha_z^\ast)}{\longrightarrow} \mathcal{O}_{Z,z}\longrightarrow 0 \end{align} $$

and Remark 3.2. Hence (3.2) and (3.3) provide

(3.5) $$ \begin{align} P_{(Y,y)}^{{(X,x)} \sqcup_{(Z,z)} {(Y,y)}}(t)=1+\left(P_{(Z,z)}^{(X,x)}(t)-1\right)P^{{(X,x)} \sqcup_{(Z,z)} {(Y,y)}}_{(X,x)}(t), \end{align} $$

and therefore

(3.6) $$ \begin{align} P^{{(X,x)} \sqcup_{(Z,z)} {(Y,y)}}_{(X,x)}(t)=\frac{1-P_{(Y,y)}^{{(X,x)} \sqcup_{(Z,z)} {(Y,y)}}(t)}{1-P^{(X,x)}_{(Z,z)}(t)}. \end{align} $$

(ii) Since the gluing ${(X,x)} \sqcup _{(Z,z)} {(Y,y)}$ is large, by definition it is also weakly large. Hence, multiplying both sides of (3.6) by $P^{(X,x)}_{(W,w)}(t)$ , one has

$$ \begin{align*}P_{(W,w)}^{{(X,x)} \sqcup_{(Z,z)} {(Y,y)}}(t) &=P^{(X,x)}_{(W,w)}(t)+P^{(X,x)}_{(Z,z)}(t)P_{(W,w)}^{{(X,x)} \sqcup_{(Z,z)} {(Y,y)}}(t) \\ &\quad -P^{{(X,x)} \sqcup_{(Z,z)} {(Y,y)}}_{(Y,y)}(t)P_{(W,w)}^{(X,x)}(t).\end{align*} $$

Therefore

$$ \begin{align*}P_{(W,w)}^{{(X,x)} \sqcup_{(Z,z)} {(Y,y)}}(t)=\frac{P_{(W,w)}^{(X,x)}(t)\left(1-P_{(Y,y)}^{{(X,x)} \sqcup_{(Z,z)} {(Y,y)}}(t)\right)}{1-P_{(Z,z)}^{(X,x)}(t)}.\end{align*} $$

As a consequence, we derive a formula to compute the Betti numbers of any subspace of the complex analytic germ $({X,x})$ as a subspace of the large gluing ${(X,x)} \sqcup _{(Z,z)} {(Y,y)}$ .

For the next two results, in order to simplify the notation, let $(\mathcal {V},v)$ denote the gluing of germs ${(X,x)} \sqcup _{(Z,z)} {(Y,y)}$ and $\beta _{i}^{T}(U):= \beta _{i}^{(T,t)}(U,u)$ , for any germs $(T,t)$ and $(U,u)$ .

Proposition 3.10 Let $(X,x)\subset (\mathbb {C}^n,x)$ , $(Y,y)\subset (\mathbb {C}^m,y)$ and $(Z,z)\subset (\mathbb {C}^l,z)$ be germs of analytic spaces. Suppose that the gluing $(\mathcal {V},v)$ is large. Then, for any $(W,w)$ subspace of $({X,x})$ ,

$$ \begin{align*}\sum_{i=0}^{j-1}\beta_i^{\mathcal{V}}(W)\beta_{j-i}^{X}({Z})=\sum_{i=0}^{j-1}\beta_i^{X}(W)\beta_{j-1}^{\mathcal{V}}{(Y)},\end{align*} $$

for each $j\geq 1$ positive integer.

Proof Since the gluing $(\mathcal {V},v)$ is large, Lemma 3.9 (ii) provides

$$ \begin{align*}P^{(\mathcal{V},v)}_{(W,w)}(t)=\frac{P^{{(X,x)}}_{(W,w)}(t)\left(1- P^{(\mathcal{V},v)}_{{(Y,y)}}(t)\right)} {1-P^{({X,x})}_{({Z,z})}(t)}.\end{align*} $$

Set $P_{(W,w)}^{({\mathcal {V},v})}(t)=\displaystyle \sum _i\beta _i^{{\mathcal {V}}}(W)t^i, P_{(W,w)}^{(X,x)}(t)=\displaystyle \sum _i\beta _i^{X}(W)t^i, P_{(Z,z)}^{({X,x})}(t)=\displaystyle \sum _i\beta _i^{X}(Z)t^i$ and $ P_{({Y,y})}^{(\mathcal {V},v)}(t)=\displaystyle \sum _i\beta _i^{{\mathcal {V}}}(Y)t^i.$ The previous equality yields

(3.7) $$ \begin{align} \sum_i\beta_i^{{\mathcal{V}}}(W)t^i\left(1-\sum_i\beta_i^{{X}}({Z})t^i\right)=\sum_i\beta_i^{{X}}(W)t^i\left(1-\sum_i\beta_i^{{\mathcal{V}}}({Y})t^i\right). \end{align} $$

Note that

$$ \begin{align*}\sum_i\beta_i^{{\mathcal{V}}}(W)t^i\left(1-\sum_i\beta_i^{{X}}(Z)t^i\right) & =\sum_i\beta_i^{{\mathcal{V}}}(W)t^i- \sum_i\beta_i^{{\mathcal{V}}}(W)t^i\sum_i\beta_i^{X}(Z)t^i \\ &= \sum_i\beta_i^{{\mathcal{V}}}(W)t^i-\sum_{j\geq 0}\left(\sum_{i=0}^{j}\beta_i^{{\mathcal{V}}}(W)\beta_{j-i}^{X}(Z)\right)t^j.\end{align*} $$

Similarly, the right side of equality (3.7) gives

(3.8) $$ \begin{align} & \sum_i\beta_i^{{\mathcal{V}}}(W)t^i-\sum_{j\geq 0}\left(\sum_{i=0}^{j}\beta_i^{{\mathcal{V}}}(W)\beta_{j-i}^{X}(Z)\right)t^j \notag \\ &\quad =\sum_i\beta_i^{{X}}(W)t^i-\sum_{j\geq 0}\left(\sum_{i=0}^{j}\beta_i^{{X}}(W)\beta_{j-i}^{{\mathcal{V}}}(Y)\right)t^j. \end{align} $$

Therefore, for each $j\geq 1$ ,

$$ \begin{align*}\beta_j^{{\mathcal{V}}}(W)-\sum_{i=0}^{j}\beta_i^{\mathcal{V}}(W)\beta_{j-i}^{{X}}(Z)=\beta_j^{X}(W)-\sum_{i=0}^{j}\beta_i^{X}(W)\beta_{j-i}^{{\mathcal{V}}}(Y).\end{align*} $$

The fact $\beta _{0}^{X}(Z)=1=\beta _0^{{\mathcal {V }}}({Y})$ furnishes

$$ \begin{align*}\sum_{i=0}^{j-1}\beta_i^{{\mathcal{V}}}(W)\beta_{j-i}^{X}(Z)=\sum_{i=0}^{j-1}\beta_i^{X}(W)\beta_{j-1}^{{\mathcal{V}}}(Y),\end{align*} $$

for each $j\geq 1$ positive integer and therefore, the desired conclusion follows.

Proof (i) By Proposition 3.10 in the case $j=1$ , one has

$$ \begin{align*}\beta_0^{\mathcal{V}}(W)\beta_1^{X}(Z)=\beta_0^{X}(W)\beta_1^{\mathcal{V}}(Y),\end{align*} $$

which implies that

$$ \begin{align*}\beta_0^{\mathcal{V}}(W)=\displaystyle\frac{{\beta_0^X(W)\beta_1^{\mathcal{V}}}(Y)}{\beta_1^{X}(Z)}.\end{align*} $$

This gives (i). The proof of (ii) and (iii) follows analogous by taking $j=2$ and $j=3$ in Proposition 3.10, respectively, together with the fact obtained in (i).

Theorem 3.13 Let ${(X,x)} \sqcup _{(Z,z)} {(Y,y)}$ be the gluing of the germs of analytic spaces $(X,x)\subset (\mathbb {C}^n,x)$ , $(Y,y)\subset (\mathbb {C}^m,y)$ and $(Z,z)\subset (\mathbb {C}^l,z)$ , satisfying one of the following conditions:

1. (i) ${(X,x)} \sqcup _{(Z,z)} {(Y,y)}$ is weakly large and there is a surjective map $\mathcal {O}_{Y,y} \twoheadrightarrow \mathcal {O}_{X,x}$ .

2. (ii) ${(X,x)} \sqcup _{(Z,z)} {(Y,y)}$ is strongly large.

If $(W,w)$ is a subspace of $(Y,y)$ , the Poincaré series of $(W,w)$ as a subspace of the gluing ${(X,x)} \sqcup _{(Z,z)} {(Y,y)}$ is given by

$$ \begin{align*}P^{{(X,x)} \sqcup_{(Z,z)} {(Y,y)}}_{(W,w)}(t)=\frac{P^{{(Y,y)}}_{(W,w)}(t)P^{(X,x)}_{{(Z,z)}}(t)} {P^{{(X,x)}}_{{(Z,z)}}(t)+P^{{(Y,y)}}_{{(Z,z)}}(t)-P^{{(X,x)}}_{{(Z,z)}}(t)P^{{(Y,z)}}_{{(Z,z)}}(t)}.\end{align*} $$

Proof (i) Note that Lemma 3.9(i) (see (3.5)) furnishes

(3.9) $$ \begin{align} P_{(Y,y)}^{{(X,x)} \sqcup_{(Z,z)} {(Y,y)}}(t)=1+P_{({Z,z})}^{({X,x})}(t)P^{{(X,x)} \sqcup_{(Z,z)} {(Y,y)}}_{({X,x})}(t)-P^{{(X,x)} \sqcup_{(Z,z)} {(Y,y)}}_{(X,x)}(t). \end{align} $$

From the exact sequence

(3.10) $$ \begin{align} 0\longrightarrow \mathrm{ker}(\beta_z^\ast)\longrightarrow \mathcal{O}_{X,x}\times_{\mathcal{O}_{Z,z}}\mathcal{O}_{Y,y}{\longrightarrow} \mathcal{O}_{X,x}\longrightarrow 0, \end{align} $$

similarly to the proof of Lemma 3.9(i), one obtains

(3.11) $$ \begin{align} P_{(X,x)}^{{(X,x)} \sqcup_{(Z,z)} {(Y,y)}}(t) & =1+tP^{{(X,x)} \sqcup_{(Z,z)} {(Y,y)}}_{(K,k)}(t) \notag \\ & = 1+ tP_{(K,k)}^{(Y,y)}(t)P_{(Y,y)}^{{(X,x)} \sqcup_{(Z,z)} {(Y,y)}}(t) \notag \\ & = 1+(P_{({Z,z})}^{({Y,y})}(t)-1)P^{{(X,x)} \sqcup_{(Z,z)} {(Y,y)}}_{({Y,y})}(t), \end{align} $$

where $(K,k)$ is the subspace of $(X,x)$ that represents the kernel of the map $\alpha _z^\ast $ , and the second equality follows by the hypothesis and Remark 3.6.

Replacing (3.11) in (3.9) one has

(3.12) $$ \begin{align} P_{{(Y,y)}}^{{(X,x)} \sqcup_{(Z,z)} {(Y,y)}}(t)=\frac{P_{(Z,z)}^{({X,x})}(t)}{P_{({Z,z})}^{({X,x})}(t)+ P_{({Z,z})}^{({Y,y})}(t)-P_{({Z,z})}^{({X,x})}(t)P_{({Z,z})}^{({Y,y})}(t)}. \end{align} $$

Again, by the hypothesis and Remark 3.6, multiplying both sides of Equation (3.12) by $P^{({Y,y})}_{(W,w)}(t)$ the desired conclusion follows.

(ii) Since ${(X,x)} \sqcup _{(Z,z)} {(Y,y)}$ is strongly large, it is also weakly large. So, as in (3.9),

(3.13) $$ \begin{align} P_{(Y,y)}^{{(X,x)} \sqcup_{(Z,z)} {(Y,y)}}(t)=1+P_{(Z,z)}^{(X,x)}(t)P^{{(X,x)} \sqcup_{(Z,z)} {(Y,y)}}_{({X,x})}(t)-P^{{(X,x)} \sqcup_{(Z,z)} {(Y,y)}}_{(X,x)}(t). \end{align} $$

With an analogous argument used in (i) and a base change, it is possible to show that

(3.14) $$ \begin{align} P_{(X,x)}^{{(X,x)} \sqcup_{(Z,z)} {(Y,y)}}(t)=1+P_{(Z,z)}^{(Y,y)}(t)P^{{(X,x)} \sqcup_{(Z,z)} {(Y,y)}}_{(Y,y)}(t)-P^{{(X,x)} \sqcup_{(Z,z)} {(Y,y)}}_{(Y,y)}(t), \end{align} $$

and therefore the statement is similarly obtained.

Corollary 3.14 Let $(X,x)\subset (\mathbb {C}^n,x)$ and $(Z,z)\subset (\mathbb {C}^l,z)$ be germs of analytic spaces. If $(W,w)$ is a subspace of $({X,x})$ , then

$$ \begin{align*}P^{{(X,x)} \sqcup_{(Z,z)} {(X,x)}}_{(W,w)}(t)=\frac{P^{{(X,x)}}_{(W,w)}(t)} {2-P^{{(X,x)}}_{(Z,z)}(t)}.\end{align*} $$

The next result shows the explicit shape of certain Betti numbers of the subspace $(W,w)$ of $(Y,y)$ seen as a subspace of the gluing ${(X,x)} \sqcup _{(Z,z)} {(Y,y)}$ . We omit the proof because it is similar to Corollary 3.12.

4 Applications

This section is devoted to show some consequences of the previous Betti numbers obtained. First, we recall some basic definitions for the convenience of the reader.

Embedding dimension: For a Noetherian local ring $(R, \mathfrak {m})$ , the minimal number of generators of $\mathfrak {m}$ will be denoted by $\operatorname {\mathrm {edim}}(R):=\dim _{\mathbb {C}}\mathfrak {m}/\mathfrak {m}^2$ and is called the embedding dimension of R. Recall that, in general, $\operatorname {\mathrm {edim}}(R) \geq \dim (R)$ . If this inequality is an equality, then R is called a regular local ring. The embedding dimension of a germ of an analytic space $(X,x)$ , denoted by $\operatorname {\mathrm {edim}}(X,x)$ , means the embedding dimension of the local ring $\mathcal {O}_{X,x}$ .

Again, let $(\mathcal {V},v)$ denote the gluing ${(X,x)} \sqcup _{(Z,z)} {(Y,y)}$ , $\beta _{i}^{T}(U):= \beta _{i}^{(T,t)}(U,u)$ , $\dim (T,t):= \dim (T)$ and the embedding dimension $\operatorname {\mathrm {edim}}(T,t):= \operatorname {\mathrm {edim}} (T)$ , for any germs $(T,t)$ and $(U,u)$ . An important fact for the rest of this section is that [Reference Ananthnarayan, Avramov and Moore2, Lemma 1.5(1.5.2)]

$$ \begin{align*}\dim (\mathcal{V})= \mathrm{max}\{\dim(X), \dim(Y)\}.\end{align*} $$

As a consequence of the characterization of the Betti numbers of the gluing of germs of complex analytic spaces (Corollary 3.12(ii) and Corollary 3.15(ii)), a formula for their embedding dimension is also provided.

Proposition 4.3 Let $(X,x)\subset (\mathbb {C}^n,x)$ , $(Y,y)\subset (\mathbb {C}^m,y)$ and $(Z,z)\subset (\mathbb {C}^l,z)$ be germs of analytic spaces such that $\dim (\mathcal {V})= \dim {(X)}.$ Suppose that the gluing $(\mathcal {V},v)$ is large. Then $(\mathcal {V},v)$ is smooth if and only if the following equality holds

$$ \begin{align*}\beta_1^{X}({Z})\left(\beta_1^{\mathcal{V}}{(Y)}{\mathrm{edim}} {(X)}-\dim {(X)}\beta_1^{{X}}({Z})+\beta_2^{\mathcal{V}}(Y)\right)=\beta_1^{\mathcal{V}}(Y)\beta_2^{{X}}({Z}).\end{align*} $$

Proof By definition, $(\mathcal {V},v)$ is smooth if and only if $\operatorname {\mathrm {edim}} (\mathcal {V})=\dim (\mathcal {V}).$ Hence, using Corollary 4.1 (i), one has that $(\mathcal {V},v)$ is smooth if and only if

(4.1) $$ \begin{align} \dim (X)=\frac{\beta_1^{\mathcal{V}}(Y)\left({\mathrm{edim}} (X)\beta_1^{X}(Z)-\beta_2^{X}(Z)\right)}{\beta_1^{X}(Z)^2}+\frac{\beta_2^{\mathcal{V}}(Y)\beta_1^{X}(Z)}{\beta_1^{X}(Z)^2}. \end{align} $$

Solving (4.1) for $\beta _1^{X}(Z)$ , one obtains that $(\mathcal {V},v)$ is smooth if and only if

$$ \begin{align*}\beta_1^{X}(Z)\left(\beta_1^{\mathcal{V}}(Y){\mathrm{edim}} (X)-\dim (X)\beta_1^{X}(Z)+\beta_2^{\mathcal{V}}(Y)\right)=\beta_1^{\mathcal{V}}(Y)\beta_2^{X}(Z),\end{align*} $$

and this shows the statement.

As an immediate consequence of Proposition 4.3, we derive the following.

Corollary 4.4 Let $(X,x)\subset (\mathbb {C}^n,x)$ , $(Y,y)\subset (\mathbb {C}^m,y)$ and $(Z,z)\subset (\mathbb {C}^l,z)$ be germs of analytic spaces such that $\dim (\mathcal {V})= \dim (X).$ Suppose that the gluing $(\mathcal {V},v)$ is large and $\beta _1^{\mathcal {V}}(Y)=\beta _1^{{X}}({Z})=1$ . Then $(\mathcal {V},v) $ is smooth if and only if

$$ \begin{align*}{\mathrm{edim}} {(X)}-\dim ({X})=\beta_2^{{X}}({Z})-\beta_2^{\mathcal{V}}({Y}).\end{align*} $$

Proposition 4.5 Let $(X,x)\subset (\mathbb {C}^n,x)$ , $(Y,y)\subset (\mathbb {C}^m,y)$ and $(Z,z)\subset (\mathbb {C}^l,z)$ be germs of analytic spaces. Suppose that $(\mathcal {V},v)$ is large and $\beta _2^X(Z)=0$ . Then $(\mathcal {V},v)$ is a complete intersection if and only if

$$ \begin{align*} & m\dim ({X})+{\mathrm{edim}} ({X})\left[\beta_2^{\mathcal{V}}(Y)\left(1-\frac{l}{m}\right)-\frac{l}{2}\right] \\ &\quad = \frac{1}{2m}\Bigl[l^2{\mathrm{edim}}({X})^2+\beta_2^{\mathcal{V}}(Y)^2\Bigr]+\frac{\beta_2^{\mathcal{V}}(Y)}{2}-\beta_3^{\mathcal{V}}(Y)-l\beta_2^X(0),\end{align*} $$

where $l=\beta _1^{\mathcal {V}}(Y)$ and $m=\beta _1^{{X}}({Z}).$

Proof By [Reference Avramov and Elias1, Theorem 7.3.3] (or [Reference Hashimoto11, Proposition 2.8.4(3)]), $(\mathcal {V},v)$ is a complete intersection if and only if

(4.2) $$ \begin{align} \beta_2^{\mathcal{V}}(0)=\binom{\beta_1^{\mathcal{V}}(0)}{2}+\beta_1^{\mathcal{V}}(0)-\dim (\mathcal{V}). \end{align} $$

Since $\beta _2^X(Z)=0$ , we obtain that the projective dimension of the germ $(Z,z)$ over $(X,x)$ is smaller than $1$ , and so $\beta _3^X(Z)=0$ . Now, by Corollary 3.12(ii)-(iii), the shape of the Betti numbers $\beta _1^{\mathcal {V}}(0)$ and $\beta _2^{\mathcal {V}}(0)$ are given by

(4.3) $$ \begin{align} \beta_1^{\mathcal{V}}(0)=\frac{\beta_1^{\mathcal{V}}(Y){\mathrm{edim}}({X})+\beta_2^{\mathcal{V}}(Y)}{\beta_1^X(Z)}; \end{align} $$

(4.4) $$ \begin{align} \beta_2^{\mathcal{V}}(0)=\frac{\beta_3^{\mathcal{V}}(Y)+{\mathrm{edim}}{(X)}\beta_2^{\mathcal{V}}(Y)+\beta_2^X(0)\beta_1^{\mathcal{V}}(Y)}{\beta_1^{X}(Z)}. \end{align} $$

Replacing (4.3) in (4.2) and comparing with (4.4), the desired result follows immediately.

It should be noted that the last results show that it is difficult to have large gluing of germs of analytic spaces that are smooth and complete intersection. For the cases of Theorem 3.13, the next result yields a better understanding of their structure.

Proof (i) Suppose that $(\mathcal {V},v)$ is smooth. Then, Corollary 4.1(ii) gives

$$ \begin{align*}\dim (Y)=\dim (\mathcal{V})= {\mathrm{edim}}(\mathcal{V})=\beta_1^{X}(Z)+{\mathrm{edim}}(Y).\end{align*} $$

So $\beta _1^{X}(Z)=0$ because $\operatorname {\mathrm {edim}}(Y)\geq \dim (Y)$ , which is a contradiction (Remark 3.11).

(ii) By (i), since $(\mathcal {V},v)$ is singular one has $\operatorname {\mathrm {edim}}(\mathcal {V})- \dim (\mathcal {V})>0$ . Hence $(\mathcal {V},v)$ is a hypersurface if and only if $\operatorname {\mathrm {edim}}(\mathcal {V})-\mathrm {depth} (\mathcal {V})=1$ . Since $(\mathcal {V},v)$ is Cohen–Macaulay by hypothesis, Corollary 4.1(ii) furnishes

$$ \begin{align*}\beta_1^{X}(Z)+{\mathrm{edim}}(Y)-\mathrm{dim} (Y)=1.\end{align*} $$

The facts $\beta _1^{X}(Z)\neq 0$ (Remark 3.11) and $\operatorname {\mathrm {edim}}(X)\geq \dim (X)$ yield that $(\mathcal {V},v)$ is a hypersurface if and only if $\beta _1^{X}(Z)=1$ and $\operatorname {\mathrm {edim}}(Y)=\mathrm {dim}(Y)$ (i.e., Y is smooth).

(iii) Set $d:=\dim (\mathcal {V})$ . By [Reference Hashimoto11, Proposition 2.8.4(3)]), $(\mathcal {V},v)$ is a complete intersection if and only if

(4.5) $$ \begin{align} \beta_2^{\mathcal{V}}(0)= \binom{\overline{e}}{2}+\overline{e}-d, \end{align} $$

where $\overline {e}:= \operatorname {\mathrm {edim}}(\mathcal {V})=\beta _1^{X}({Z})+\operatorname {\mathrm {edim}}({Y})$ (Corollary 4.1(ii)). By Corollary 3.15(iii) and the fact that $(X,x)$ is a complete intersection [Reference Hashimoto11, Proposition 2.8.4(3)] yield

(4.6) $$ \begin{align}\beta_2^{\mathcal{V}}(0)= \beta_1^{Y}(Z)\beta_1^{{X}}({Z})+\beta_2^{X}(Z)+e_2\beta_1^{X}(Z)+\binom{e_2}{2}+e_2-d,\end{align} $$

where $e_2:=\operatorname {\mathrm {edim}}(Y)$ . Therefore, comparing (4.5) and (4.6) we obtains that $(\mathcal {V},v)$ is a complete intersection if and only if

$$ \begin{align*}\beta_1^{X}(Z)^2+\beta_1^{X}(Z)=2\left(\beta_1^{Y}(Z)\beta_1^{X}(Z)+\beta_2^{X}(Z)\right).\end{align*} $$

The desired conclusion follows, because $\beta _1^{Y}(Z)\neq 0\neq \beta _1^{X}(Z)$ (Remark 3.11).

(iv) Suppose that $(Y,y)$ is smooth. Since $\beta _1^{X}(Z)\neq 0$ (Remark 3.11), the hypothesis and Corollary 3.15(i) and (ii) provide $\beta _1^{X}(Z)=1$ and $\beta _1^{Y}(W)=0$ . Therefore $(\mathcal {V},v)$ is Gorenstein by (ii). The converse immediately follows from [Reference Endo, Goto and Isobe5, Proposition 4.19].

Example 4.8 Let $X=V(x^5)$ , $Y=V(y^5)$ and $Z=V(z^2)$ be analytic subspaces of $\mathbb {C}$ and consider $(X,0)$ , $(Y,0)$ and $(Z,0)$ their respective germs at the origin. Note that $(X,0)$ is a complete intersection and the gluing $(X,0)\sqcup _{(Z,0)}(Y,0)$ is defined by the ideal

$$ \begin{align*}\mathcal{I}_{(X,0)\sqcup_{(Z,0)}(Y,0)}=(u^ 5,uv^2,v^2-u^2v)\end{align*} $$

in $\mathbb {C}\{u,v\}$ , which is not a complete intersection [Reference Ananthnarayan, Avramov and Moore2, Example 3.4]. By Example 3.5(ii), the gluing $(X,0)\sqcup _{(Z,0)}(Y,0)$ is large, because the ideal $(y^2)$ in $\mathbb {C}\{y\}/(y^5)$ is the kernel of the map $\mathbb {C}\{y\}/(y^5)\to \mathbb {C}\{z\}/(z^2)$ and it is a weak complete intersection ideal (by [Reference Rahmati, Striuli and Yang18, Example 2.3(ii)]). Since $\beta _1^Y(Z)=\beta _1^X(Z)=\beta _2^X(Z)=1$ , one has

$$ \begin{align*}\frac{\beta_1^X(Z)^2+\beta_1^X(Z)}{\beta_1^X(Z)\beta_1^Y(Z)+\beta_2^X(Z)}\neq 2.\end{align*} $$

As mentioned in Remark 3.6, the gluing ${(X,x)} \sqcup _{(Z,z)} {(X,x)}$ is always strongly large. Since the dimension of $\dim {(X,x)} \sqcup _{(Z,z)} {(X,x)}$ and $\dim (X)$ are equal, as a consequence of Theorem 3.13 we derive the following result.

Proof The proof of (i), (ii) and (iv) are immediate consequences of Theorem 3.13(i)-(ii)-(iv). For (iii), Theorem 3.13(iii) furnishes

(4.7) $$ \begin{align} \beta_1^X(Z)-\beta_1^X(Z)^2=2\beta_2^X(Z).\end{align} $$

Note that, if $\beta _1^X(Z)>1$ , then left side of (4.7) is a negative number. Since $\beta _2^X(Z)\geq 0$ , the equality (4.7) occurs if and only if $\beta _1^X(Z)=1$ and $\beta _2^X(Z)=0$ or $\beta _1^X(Z)=0$ and $\beta _2^X(Z)=0$ . But $\beta _1^X(Z)\neq 0$ (Remark 3.11), and therefore the result follows.