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Phil Tosteson
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One way to think about why this is hard is to consider the relationship between $C_*(\tilde X)$ and $C_*(X)$.

The group $G := \pi_1 X$ acts on $C_*(\tilde X)$ and $C_*(X)$ is quasi-isomorphic to the space of coinvariants, i.e. $H_i(X) = H_i(C_*(X)/G)$. Thus an analogous question is: given a group $G$ and and complex $D \in Ch(Ab^G)$, can we recover $D$ from $D_G$?

The answer to this question is clearly no. For instance, if $G$ is finite, and $D$ a non-trivial irreducible $G$ representation, we have $D_G = 0$. In general, the most you can hope to recover are the "unipotent" $G$ representations-- i.e. those which are a finite extension of direct sums of trivial representations. On the subcategory of chain complexes whose homology is unipotent, coinvariant functor is conservative in a derived.sense sense So you can hope to recover $D$ from $D_G$, together with some extra algebraic structure.

This is why Eilenberg-Moore spectral sequence will only converge to what you want when $\pi_1 X$ acts unipotently on $H_*(\tilde X)$. In general, it will converge to something different--which we might call the homology of the unipotent completion of $C_*(\tilde X)$.

One way to think about why this is hard is to consider the relationship between $C_*(\tilde X)$ and $C_*(X)$.

The group $G := \pi_1 X$ acts on $C_*(\tilde X)$ and $C_*(X)$ is quasi-isomorphic to the space of coinvariants, i.e. $H_i(X) = H_i(C_*(X)/G)$. Thus an analogous question is: given a group $G$ and and complex $D \in Ch(Ab^G)$, can we recover $D$ from $D_G$?

The answer to this question is clearly no. For instance, if $G$ is finite, and $D$ a non-trivial irreducible $G$ representation, we have $D_G = 0$. In general, the most you can hope to recover are the "unipotent" $G$ representations-- i.e. those which are a finite extension of direct sums of trivial representations. On the subcategory of chain complexes whose homology is unipotent, coinvariant functor is conservative in a derived.sense So you can hope to recover $D$ from $D_G$, together with some extra algebraic structure.

This is why Eilenberg-Moore spectral sequence will only converge to what you want when $\pi_1 X$ acts unipotently on $H_*(\tilde X)$. In general, it will converge to something different--which we might call the homology of the unipotent completion of $C_*(\tilde X)$.

One way to think about why this is hard is to consider the relationship between $C_*(\tilde X)$ and $C_*(X)$.

The group $G := \pi_1 X$ acts on $C_*(\tilde X)$ and $C_*(X)$ is quasi-isomorphic to the space of coinvariants, i.e. $H_i(X) = H_i(C_*(X)/G)$. Thus an analogous question is: given a group $G$ and and complex $D \in Ch(Ab^G)$, can we recover $D$ from $D_G$?

The answer to this question is clearly no. For instance, if $G$ is finite, and $D$ a non-trivial irreducible $G$ representation, we have $D_G = 0$. In general, the most you can hope to recover are the "unipotent" $G$ representations-- i.e. those which are a finite extension of direct sums of trivial representations. On the subcategory of chain complexes whose homology is unipotent, coinvariant functor is conservative in a derived sense So you can hope to recover $D$ from $D_G$, together with some extra algebraic structure.

This is why Eilenberg-Moore spectral sequence will only converge to what you want when $\pi_1 X$ acts unipotently on $H_*(\tilde X)$. In general, it will converge to something different--which we might call the homology of the unipotent completion of $C_*(\tilde X)$.

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Phil Tosteson
  • 3.9k
  • 1
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One way to think about why this is hard is to consider the relationship between $C_*(\tilde X)$ and $C_*(X)$.

The group $G := \pi_1 X$ acts on $C_*(\tilde X)$ and $C_*(X)$ is quasi-isomorphic to the space of coinvariants, i.e. $H_i(X) = H_i(C_*(X)/G)$. Thus an analogous question is: given a group $G$ and and complex $D \in Ch(Ab^G)$, can we recover $D$ from $D_G$?

The answer to this question is clearly no. For instance, if $G$ is finite, and $D$ a non-trivial irreducible $G$ representation, we have $D_G = 0$. In general, the most you can hope to recover are the "unipotent" $G$ representations-- i.e. those which are a finite extension of direct sums of trivial representations. On the subcategory of chain complexes whose homology is unipotent representations, the coinvariant functor is conservative (a variant of Nakayama's lemma)in a derived.sense So you can hope to recover $D$ from $D_G$, together with some extra algebraic structure.

This is why Eilenberg-Moore spectral sequence will only converge to what you want when $\pi_1 X$ acts unipotently on $H_*(\tilde X)$. In general, it will converge to something different--which we might call the homology of the unipotent completion of $C_*(\tilde X)$.

One way to think about why this is hard is to consider the relationship between $C_*(\tilde X)$ and $C_*(X)$.

The group $G := \pi_1 X$ acts on $C_*(\tilde X)$ and $C_*(X)$ is quasi-isomorphic to the space of coinvariants, i.e. $H_i(X) = H_i(C_*(X)/G)$. Thus an analogous question is: given a group $G$ and and complex $D \in Ch(Ab^G)$, can we recover $D$ from $D_G$?

The answer to this question is clearly no. For instance, if $G$ is finite, and $D$ a non-trivial irreducible $G$ representation, we have $D_G = 0$. In general, the most you can hope to recover are the "unipotent" $G$ representations-- i.e. those which are a finite extension of direct sums trivial representations. On the subcategory of unipotent representations, the coinvariant functor is conservative (a variant of Nakayama's lemma). So you can hope to recover $D$ from $D_G$, together with some extra algebraic structure.

This is why Eilenberg-Moore spectral sequence will only converge to what you want when $\pi_1 X$ acts unipotently on $H_*(\tilde X)$. In general, it will converge to something different--which we might call the homology of the unipotent completion of $C_*(\tilde X)$.

One way to think about why this is hard is to consider the relationship between $C_*(\tilde X)$ and $C_*(X)$.

The group $G := \pi_1 X$ acts on $C_*(\tilde X)$ and $C_*(X)$ is quasi-isomorphic to the space of coinvariants, i.e. $H_i(X) = H_i(C_*(X)/G)$. Thus an analogous question is: given a group $G$ and and complex $D \in Ch(Ab^G)$, can we recover $D$ from $D_G$?

The answer to this question is clearly no. For instance, if $G$ is finite, and $D$ a non-trivial irreducible $G$ representation, we have $D_G = 0$. In general, the most you can hope to recover are the "unipotent" $G$ representations-- i.e. those which are a finite extension of direct sums of trivial representations. On the subcategory of chain complexes whose homology is unipotent, coinvariant functor is conservative in a derived.sense So you can hope to recover $D$ from $D_G$, together with some extra algebraic structure.

This is why Eilenberg-Moore spectral sequence will only converge to what you want when $\pi_1 X$ acts unipotently on $H_*(\tilde X)$. In general, it will converge to something different--which we might call the homology of the unipotent completion of $C_*(\tilde X)$.

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Phil Tosteson
  • 3.9k
  • 1
  • 19
  • 25

One way to think about why this is hard is to consider the relationship between $C_*(\tilde X)$ and $C_*(X)$.

The group $G := \pi_1 X$ acts on $C_*(\tilde X)$ and $C_*(X)$ is quasi-isomorphic to the space of coinvariants, i.e. $H_i(X) = H_i(C_*(X)/G)$. Thus an analogous question is: given a group $G$ and and complex $D \in Ch(Ab^G)$, can we recover $D$ from $D_G$?

The answer to this question is clearly no. For instance, if $G$ is finite, and $D$ a non-trivial irreducible $G$ representation, we have $D_G = 0$. In general, the most you can hope to recover are the "unipotent" $G$ representations-- i.e. those which are a finite extension of direct sums trivial representations. On the subcategory of unipotent representations, the coinvariant functor is conservative (a variant of Nakayama's lemma). So you can hope to recover $D$ from $D_G$, together with some extra algebraic structure.

This is why Eilenberg-Moore spectral sequence will only converge to what you want when $\pi_1 X$ acts unipotently on $H_*(\tilde X)$. In general, it will converge to something different--which we might call the homology of the unipotent completion of $C_*(\tilde X)$.