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Elías Guisado Villalgordo
Elías Guisado Villalgordo
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$\def\M{\mathcal{M}} \def\ch{\operatorname{Ch}}$Let $\mathcal{A}$ be an $A_\infty$-category. An $A_\infty$-module $\M$ over $\mathcal{A}$ is an $A_\infty$-functor $\mathcal{A}^\mathrm{op}\to\operatorname{Ch}$, where $\ch$ is the dg-category of chain complexes of vector spaces [S, (1j)]. In ibid. we find:

(3e) Cones. Let $Y_0, Y_1$ be objects of $\mathcal{A}$, and $c \in \operatorname{hom}_{\mathcal{A}}^0\left(Y_0, Y_1\right)$ a degree zero cocycle $\left(\mu_{\mathcal{A}}^1(c)=0\right)$. The abstract mapping cone of $c$ is the $A_{\infty}$-module $\mathcal{C}=\mathcal{C}one(c)$ defined by $$ \mathcal{C}(X)=\operatorname{hom}_{\mathcal{A}}\left(X, Y_0\right)[1] \oplus \operatorname{hom}_{\mathcal{A}}\left(X, Y_1\right), $$ $$ \begin{gathered} \mu_{\mathcal{C}}^d\left(\left(b_0, b_1\right), a_{d-1}, \ldots, a_1\right)=\left(\mu_{\mathcal{A}}^d\left(b_0, a_{d-1}, \ldots, a_1\right), \mu_{\mathcal{A}}^d\left(b_1, a_{d-1}, \ldots, a_1\right)+\right. \\ \left.+\mu_{\mathcal{A}}^{d+1}\left(c, b_0, a_{d-1}, \ldots, a_1\right)\right). \end{gathered} $$

A straightforward computation shows that (1.39) is satisfied.

Where (1.39) is the $A_\infty$-law for an $A_\infty$-module.

In [S, Remark 3.5], it is claimed that if $\mathcal{A}$ is c-unital (meaning $H(\mathcal{A})$ is unital), and $e_Z\in\operatorname{Hom}_\mathcal{A}(Z,Z)$ is a cocycle representing the identity of $Z$ in $H(\mathcal{A})$, then $H(\mathcal{C}one(e_Z))=0$ (one needs this in a step of the proof of [S, Proposition 3.14], that says that if $\mathcal{A}$ is an $A_\infty$-triangulated category, then $H(\mathcal{A})$ is a triangulated category; equation $H(\mathcal{C}one(e_Z))=0$ means that $H(\mathcal{A})$ has a zero object).

However, I am unable to show that $\mathcal{C}one(e_Z)$ has vanishing cohomology. Homogeneous cocycles of $\mathcal{C}(X)$ are of the form $(b_0,b_1)=((-1)^{|b_1|}\mu^1_\mathcal{A}(b_1+h),b_1)$, where $\mu^2_\mathcal{A}(e_Y,b_0)=(-1)^{|b_0|}b_0+\mu_\mathcal{A}^1h$. On the other hand, equation $(b_0,b_1)=\mu^1_\mathcal{C}((c_0,c_1))$ means \begin{align*} (-1)^{|b_1|}\mu^1_\mathcal{A}(b_1+h)&=\mu_\mathcal{A}^1(c_0)\\ b_1&=\mu_\mathcal{A}^1(c_1)+\mu_\mathcal{A}^2(e_Y,c_0) \end{align*} Here's where I'm stuck. How can one find such $(c_0,c_1)$?

I know show the result if $\mathcal{A}$ is strictly unital and $e_Z$ is a strict identity. I think one can then reduce the problem from the c-unital case to the strictly unital one by [S, Proposition 3.14, Corollary 2.13]. However, this seems to be an overkill.

References

[S] P. Seidel, Fukaya Categories and Picard-Lefschetz Theory

$\def\M{\mathcal{M}} \def\ch{\operatorname{Ch}}$Let $\mathcal{A}$ be an $A_\infty$-category. An $A_\infty$-module $\M$ over $\mathcal{A}$ is an $A_\infty$-functor $\mathcal{A}^\mathrm{op}\to\operatorname{Ch}$, where $\ch$ is the dg-category of chain complexes of vector spaces [S, (1j)]. In ibid. we find:

(3e) Cones. Let $Y_0, Y_1$ be objects of $\mathcal{A}$, and $c \in \operatorname{hom}_{\mathcal{A}}^0\left(Y_0, Y_1\right)$ a degree zero cocycle $\left(\mu_{\mathcal{A}}^1(c)=0\right)$. The abstract mapping cone of $c$ is the $A_{\infty}$-module $\mathcal{C}=\mathcal{C}one(c)$ defined by $$ \mathcal{C}(X)=\operatorname{hom}_{\mathcal{A}}\left(X, Y_0\right)[1] \oplus \operatorname{hom}_{\mathcal{A}}\left(X, Y_1\right), $$ $$ \begin{gathered} \mu_{\mathcal{C}}^d\left(\left(b_0, b_1\right), a_{d-1}, \ldots, a_1\right)=\left(\mu_{\mathcal{A}}^d\left(b_0, a_{d-1}, \ldots, a_1\right), \mu_{\mathcal{A}}^d\left(b_1, a_{d-1}, \ldots, a_1\right)+\right. \\ \left.+\mu_{\mathcal{A}}^{d+1}\left(c, b_0, a_{d-1}, \ldots, a_1\right)\right). \end{gathered} $$

A straightforward computation shows that (1.39) is satisfied.

Where (1.39) is the $A_\infty$-law for an $A_\infty$-module.

In [S, Remark 3.5], it is claimed that if $\mathcal{A}$ is c-unital (meaning $H(\mathcal{A})$ is unital), and $e_Z\in\operatorname{Hom}_\mathcal{A}(Z,Z)$ is a cocycle representing the identity of $Z$ in $H(\mathcal{A})$, then $H(\mathcal{C}one(e_Z))=0$ (one needs this in a step of the proof of [S, Proposition 3.14], that says that if $\mathcal{A}$ is an $A_\infty$-triangulated category, then $H(\mathcal{A})$ is a triangulated category; equation $H(\mathcal{C}one(e_Z))=0$ means that $H(\mathcal{A})$ has a zero object).

However, I am unable to show that $\mathcal{C}one(e_Z)$ has vanishing cohomology. Homogeneous cocycles of $\mathcal{C}(X)$ are of the form $(b_0,b_1)=((-1)^{|b_1|}\mu^1_\mathcal{A}(b_1+h),b_1)$, where $\mu^2_\mathcal{A}(e_Y,b_0)=(-1)^{|b_0|}b_0+\mu_\mathcal{A}^1h$. On the other hand, equation $(b_0,b_1)=\mu^1_\mathcal{C}((c_0,c_1))$ means \begin{align*} (-1)^{|b_1|}\mu^1_\mathcal{A}(b_1+h)&=\mu_\mathcal{A}^1(c_0)\\ b_1&=\mu_\mathcal{A}^1(c_1)+\mu_\mathcal{A}^2(e_Y,c_0) \end{align*} Here's where I'm stuck. How can one find such $(c_0,c_1)$?

References

[S] P. Seidel, Fukaya Categories and Picard-Lefschetz Theory

$\def\M{\mathcal{M}} \def\ch{\operatorname{Ch}}$Let $\mathcal{A}$ be an $A_\infty$-category. An $A_\infty$-module $\M$ over $\mathcal{A}$ is an $A_\infty$-functor $\mathcal{A}^\mathrm{op}\to\operatorname{Ch}$, where $\ch$ is the dg-category of chain complexes of vector spaces [S, (1j)]. In ibid. we find:

(3e) Cones. Let $Y_0, Y_1$ be objects of $\mathcal{A}$, and $c \in \operatorname{hom}_{\mathcal{A}}^0\left(Y_0, Y_1\right)$ a degree zero cocycle $\left(\mu_{\mathcal{A}}^1(c)=0\right)$. The abstract mapping cone of $c$ is the $A_{\infty}$-module $\mathcal{C}=\mathcal{C}one(c)$ defined by $$ \mathcal{C}(X)=\operatorname{hom}_{\mathcal{A}}\left(X, Y_0\right)[1] \oplus \operatorname{hom}_{\mathcal{A}}\left(X, Y_1\right), $$ $$ \begin{gathered} \mu_{\mathcal{C}}^d\left(\left(b_0, b_1\right), a_{d-1}, \ldots, a_1\right)=\left(\mu_{\mathcal{A}}^d\left(b_0, a_{d-1}, \ldots, a_1\right), \mu_{\mathcal{A}}^d\left(b_1, a_{d-1}, \ldots, a_1\right)+\right. \\ \left.+\mu_{\mathcal{A}}^{d+1}\left(c, b_0, a_{d-1}, \ldots, a_1\right)\right). \end{gathered} $$

A straightforward computation shows that (1.39) is satisfied.

Where (1.39) is the $A_\infty$-law for an $A_\infty$-module.

In [S, Remark 3.5], it is claimed that if $\mathcal{A}$ is c-unital (meaning $H(\mathcal{A})$ is unital), and $e_Z\in\operatorname{Hom}_\mathcal{A}(Z,Z)$ is a cocycle representing the identity of $Z$ in $H(\mathcal{A})$, then $H(\mathcal{C}one(e_Z))=0$ (one needs this in a step of the proof of [S, Proposition 3.14], that says that if $\mathcal{A}$ is an $A_\infty$-triangulated category, then $H(\mathcal{A})$ is a triangulated category; equation $H(\mathcal{C}one(e_Z))=0$ means that $H(\mathcal{A})$ has a zero object).

However, I am unable to show that $\mathcal{C}one(e_Z)$ has vanishing cohomology. Homogeneous cocycles of $\mathcal{C}(X)$ are of the form $(b_0,b_1)=((-1)^{|b_1|}\mu^1_\mathcal{A}(b_1+h),b_1)$, where $\mu^2_\mathcal{A}(e_Y,b_0)=(-1)^{|b_0|}b_0+\mu_\mathcal{A}^1h$. On the other hand, equation $(b_0,b_1)=\mu^1_\mathcal{C}((c_0,c_1))$ means \begin{align*} (-1)^{|b_1|}\mu^1_\mathcal{A}(b_1+h)&=\mu_\mathcal{A}^1(c_0)\\ b_1&=\mu_\mathcal{A}^1(c_1)+\mu_\mathcal{A}^2(e_Y,c_0) \end{align*} Here's where I'm stuck. How can one find such $(c_0,c_1)$?

I know show the result if $\mathcal{A}$ is strictly unital and $e_Z$ is a strict identity. I think one can then reduce the problem from the c-unital case to the strictly unital one by [S, Proposition 3.14, Corollary 2.13]. However, this seems to be an overkill.

References

[S] P. Seidel, Fukaya Categories and Picard-Lefschetz Theory

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Elías Guisado Villalgordo
Elías Guisado Villalgordo
  • 1.3k
  • 6
  • 19

The cone of the c-identity of an $A_\infty$-module has zero cohomology

$\def\M{\mathcal{M}} \def\ch{\operatorname{Ch}}$Let $\mathcal{A}$ be an $A_\infty$-category. An $A_\infty$-module $\M$ over $\mathcal{A}$ is an $A_\infty$-functor $\mathcal{A}^\mathrm{op}\to\operatorname{Ch}$, where $\ch$ is the dg-category of chain complexes of vector spaces [S, (1j)]. In ibid. we find:

(3e) Cones. Let $Y_0, Y_1$ be objects of $\mathcal{A}$, and $c \in \operatorname{hom}_{\mathcal{A}}^0\left(Y_0, Y_1\right)$ a degree zero cocycle $\left(\mu_{\mathcal{A}}^1(c)=0\right)$. The abstract mapping cone of $c$ is the $A_{\infty}$-module $\mathcal{C}=\mathcal{C}one(c)$ defined by $$ \mathcal{C}(X)=\operatorname{hom}_{\mathcal{A}}\left(X, Y_0\right)[1] \oplus \operatorname{hom}_{\mathcal{A}}\left(X, Y_1\right), $$ $$ \begin{gathered} \mu_{\mathcal{C}}^d\left(\left(b_0, b_1\right), a_{d-1}, \ldots, a_1\right)=\left(\mu_{\mathcal{A}}^d\left(b_0, a_{d-1}, \ldots, a_1\right), \mu_{\mathcal{A}}^d\left(b_1, a_{d-1}, \ldots, a_1\right)+\right. \\ \left.+\mu_{\mathcal{A}}^{d+1}\left(c, b_0, a_{d-1}, \ldots, a_1\right)\right). \end{gathered} $$

A straightforward computation shows that (1.39) is satisfied.

Where (1.39) is the $A_\infty$-law for an $A_\infty$-module.

In [S, Remark 3.5], it is claimed that if $\mathcal{A}$ is c-unital (meaning $H(\mathcal{A})$ is unital), and $e_Z\in\operatorname{Hom}_\mathcal{A}(Z,Z)$ is a cocycle representing the identity of $Z$ in $H(\mathcal{A})$, then $H(\mathcal{C}one(e_Z))=0$ (one needs this in a step of the proof of [S, Proposition 3.14], that says that if $\mathcal{A}$ is an $A_\infty$-triangulated category, then $H(\mathcal{A})$ is a triangulated category; equation $H(\mathcal{C}one(e_Z))=0$ means that $H(\mathcal{A})$ has a zero object).

However, I am unable to show that $\mathcal{C}one(e_Z)$ has vanishing cohomology. Homogeneous cocycles of $\mathcal{C}(X)$ are of the form $(b_0,b_1)=((-1)^{|b_1|}\mu^1_\mathcal{A}(b_1+h),b_1)$, where $\mu^2_\mathcal{A}(e_Y,b_0)=(-1)^{|b_0|}b_0+\mu_\mathcal{A}^1h$. On the other hand, equation $(b_0,b_1)=\mu^1_\mathcal{C}((c_0,c_1))$ means \begin{align*} (-1)^{|b_1|}\mu^1_\mathcal{A}(b_1+h)&=\mu_\mathcal{A}^1(c_0)\\ b_1&=\mu_\mathcal{A}^1(c_1)+\mu_\mathcal{A}^2(e_Y,c_0) \end{align*} Here's where I'm stuck. How can one find such $(c_0,c_1)$?

References

[S] P. Seidel, Fukaya Categories and Picard-Lefschetz Theory