Let $C$ be a connected filtered coalgebra over a field $k$. Maybe $k$ has characteristic $0$ (though I don't know where this can be of use). Let $1$ denote the unique element of $C\_0$ mapping to $1\in k$ under the counit. Let $\overline{C^{\ast}}$ be the ideal of the dual algebra $C^{\ast}$ which consists of all $f\in C^{\ast}$ such that $f\left(1\right)=0$.


Let some $f\in \overline{C^{\ast}}$ map every primitive element of $C$ to $0$. Does this yield that $f\in \overline{C^{\ast}}^2$?


Note that the converse of this (every element of $\overline{C^{\ast}}^2$ maps every primitive element of $C$ to $0$) is obviously true. Also note that this is true if $C$ is finite-dimensional. In fact, it is then a consequence (by dualization) of the fact that if $\left(A,\varepsilon\right)$ is an augmented $k$-algebra (not necessarily finite-dimensional), and $A^+$ is its augmentation ideal, then there is a canonical isomorphism between the $k$-vector space $\left(A^+ / \left(A^+\right)^2\right)^{\ast}$ and the $k$-vector space of $\left(\varepsilon,\varepsilon\right)$-derivations $A\to k$.



Linear functional kills all primitives of a connected filtered coalgebra => it lies in m^2?
Okay, I should have thought a bit more about this question before posting it on MathOverflow: the answer is No, for a combination of rather straightforward reasons. Moreover, the answer is No even for a locally finite connected graded coalgebra over a field $k$, where "locally finite" means that every graded component is finite-dimensional. For the sake of completeness (and to avoid [this](http://xkcd.com/979/)), here is how to construct a counterexample:


**§1.** Let $C$ be a coalgebra. Let $f\in\overline{C^{\ast}}^{2}$ be arbitrary. Then, the *rank* of $f$ will mean the smallest $m\in \mathbb{N}$ such that $f$ can be written in the form $f=\sum\limits\_{i=1} ^{m}\lambda\_{i}g\_{i}h\_{i}$ with $\lambda\_{i}\in k$, $g\_{i}\in\overline {C^{\ast}}$ and $h\_{i}\in\overline{C^{\ast}}$.


**§2.** Let $k=\mathbb{R}$. Let $n\in\mathbb{N}$ be positive. Let $T$ denote the graded $k$-bialgebra $k\left[ X\_{1},X\_{2},...,X\_{n}\right] $ which, as a $k$-algebra, is the polynomial ring $k\left[ X\_{1},X\_{2},...,X\_{n}\right] $, and whose coalgebra structure is given by $\Delta\left( X\_{i}\right) =X\_{i}\otimes1+1\otimes X\_{i}$ and $\varepsilon\left( X\_{i}\right) =0$ for all $i$. This is clearly a locally finite connected graded $k$-bialgebra.


Let $f\_{n}:T\rightarrow k$ be the $k$-linear map which sends every polynomial to the sum of its coefficients before $X\_{i}X\_{j}$ over all $1\leq i < j\leq n$. Then, this map $f\_{n}$ lies in $\overline{T^{\ast}}^{2}$ and has rank $n-1$.


*Proof.* The element $f\_{n}$ of $T^{\ast}$ can be written as $f\_{n} =\sum\limits\_{1\leq i < j\leq n}\xi\_{i}\xi\_{j}$, where $\xi\_{\ell}:T\rightarrow k$ is the $k$-linear map which sends every polynomial to its coefficient before $X\_{\ell}$.


From $f\_{n}=\sum\limits\_{1\leq i < j\leq n}\xi\_{i}\xi\_{j}=\sum\limits\_{i=1} ^{n-1}\underbrace{\xi\_{i}}\_{\in\overline{T^{\ast}}}\underbrace{\left( \xi\_{i+1}+\xi\_{i+2}+...+\xi\_{n}\right) }\_{\in\overline{T^{\ast}}}$, it is clear that $f\_{n}$ lies in $\overline{T^{\ast}}^{2}$ and has rank $\leq n-1$. Now we are going to prove that it has rank $\geq n-1$.


Let $\pi\_{1}:T\rightarrow T\_{1}$ be the projection of the $k$-vector space $T=T\_{0}\oplus T\_{1}\oplus T\_{2}\oplus...$ onto its $1$st graded component $T\_{1}$. Let $\iota\_{1}:T\_{1}\rightarrow T$ be the canonical inclusion. Then, the map $\iota\_{1}\circ\pi\_{1}:T\rightarrow T$ sends every element of $T$ to its first graded component (viewed as an element of $T$).


Notice that for every $\left( i\_{1},i\_{2},...,i\_{n}\right) \in\mathbb{N} ^{n}$, the map $\xi\_{1}^{i\_{1}}\xi\_{2}^{i\_{2}}...\xi\_{n}^{i\_{n}}$ sends every polynomial to its coefficient before $X\_{1}^{i\_{1}}X\_{2}^{i\_{2}} ...X\_{n}^{i\_{n}}$. Hence, the elements $\xi\_{1}$, $\xi\_{2}$, $...$, $\xi\_{n}$ of $T^{\ast}$ are algebraically independent. Let $L$ be the $k$-subspace of $T^{\ast}$ they generate. Then, it is clear that $L$ consists of all $k$-linear maps $\phi:T\rightarrow k$ which are zero on all but the $1$st graded components of $T$. Hence, for every map $\phi\in T^{\ast}$, the map $\phi\circ\iota\_{1}\circ\pi\_{1}$ belongs to $L$.


Now, assume that $f\_{n}$ has rank $ < n-1$. Then, we can write $f\_{n}$ in the form $f\_{n}=\sum\limits\_{i=1}^{m}\lambda\_{i}g\_{i}h\_{i}$ with $\lambda\_{i}\in k$, $g\_{i}\in\overline{T^{\ast}}$ and $h\_{i}\in\overline{T^{\ast}}$ for some $m < n-1$. Now, it is easy to see that the equation $f\_{n}=\sum\limits\_{i=1} ^{m}\lambda\_{i}g\_{i}h\_{i}$ remains true if we replace all $g\_{i}$ by $g\_{i}\circ\iota\_{1}\circ\pi\_{1}$ and replace all $h\_{i}$ by $h\_{i}\circ \iota\_{1}\circ\pi\_{1}$. (In fact, to prove that it remains true, just check it on every graded component of $T$ separately, using the gradedness of the comultiplication and the fact that $g\_{i}$ and $h\_{i}$ are zero on $T\_{0}$.) But this replacement has the effect that the *new* $g\_{i}$ and $h\_{i}$ belong to $L$ (because for every map $\phi\in T^{\ast}$, the map $\phi\circ\iota \_{1}\circ\pi\_{1}$ belongs to $L$).


Thus we have written $f\_{n}$ in the form $f\_{n}=\sum\limits\_{i=1}^{m} \lambda\_{i}g\_{i}h\_{i}$ with $\lambda\_{i}\in k$, $g\_{i}\in L$ and $h\_{i}\in L$ for some $m < n-1$. Since $\xi\_{1}$, $\xi\_{2}$, $...$, $\xi\_{n}$ are algebraically independent, we can view them as polynomial indeterminates, and then the equation $f\_{n}=\sum\limits\_{i=1}^{m}\lambda\_{i}g\_{i}h\_{i}$ becomes a
polynomial identity between the polynomial $f\_{n}=\sum\limits\_{1\leq i < j\leq n}\xi\_{i}\xi\_{j}$ and the polynomial $\sum\limits\_{i=1}^{m}\lambda\_{i} g\_{i}h\_{i}$, where $g\_{i}$ and $h\_{i}$ are homogeneous polynomials of degree $1$ in $\xi\_{1}$, $\xi\_{2}$, $...$, $\xi\_{n}$ (because they lie in $L=\left\langle \xi\_{1},\xi\_{2},...,\xi\_{n}\right\rangle $). Now, since homogeneous polynomials of degree $2$ are quadratic forms, and homogeneous polynomials of degree $1$ are linear forms, we can rewrite this result in a linear-algebraic way: The quadratic form $\sum\limits\_{1\leq i < j\leq n}\xi \_{i}\xi\_{j}$ on the vector space $k^{n}$ can be written as a $k$-linear combination of $m$ products of linear forms $g\_{i}$ and $h\_{i}$. As a consequence, of course, the quadratic form $\sum\limits\_{1\leq i < j\leq n} \xi\_{i}\xi\_{j}$ must be zero on the intersection $\bigcap\limits\_{i=1} ^{m}\operatorname\*{Ker}g\_{i}$. Since $\bigcap\limits\_{i=1}^{m}
\operatorname\*{Ker}g\_{i}$ is a subspace of $k^{n}$ of dimension at least $2$ (because $m < n-1$), this yields that the quadratic form $\sum\limits\_{1\leq i < j\leq n}\xi\_{i}\xi\_{j}$ is zero on some subspace of $k^{n}$ of dimension at least $2$. On the other hand, since $k=\mathbb{R}$, we know that the quadratic form $\sum\limits\_{1\leq i < j\leq n}\xi\_{i}\xi\_{j}$ is negative definite on some subspace of $k^{n}$ of dimension $n-1$ (see, for instance, Theorem 2 in <http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2429180#p2429180> ). The intersection of these two subspaces is nonzero (since $2+\left( n-1\right) >n$), and on this intersection we get a contradiction. This falsifies our assumption that $f\_{n}$ have rank $ < n-1$. Thus, $f\_{n}$ must have rank exactly $n-1$, and §2 is proven.


[Note that we used $k=\mathbb{R}$ here, but for an arbitrary $k$ we could still have proven that $f\_{n}$ has rank between $\dfrac{n}{2}$ and $n-1$, and this would still be enough for the below argument once some minor changes are made.]


**§3.** Let $C$ be a graded coalgebra. Let $m\in\mathbb{N}$ be positive. Then, we define a graded coalgebra $C\_{\left( m\right) }$ as the coalgebra $C$ with the following new grading: $\left( C\_{\left( m\right) }\right) \_{i}=\left\{ \begin{array} {c} C\_{i/m}\text{, if }m\mid i;\\ 0\text{, if }m\nmid i \end{array} \right. $ for all $i\in\mathbb{N}$.


It is clear that this new graded coalgebra $C\_{\left( m\right) }$ is well-defined, and if $C$ is connected, then $C\_{\left( m\right) }$ is also connected.


**§4.** Some notations: For every connected graded coalgebra $C$, let $C^{+}$ denote the positive part of $C$ (that is, the subspace $C\_{1}\oplus C\_{2}\oplus C\_{3}\oplus...=\operatorname\*{Ker}\varepsilon$), and for every $c\in C^{+}$, we denote the element $\Delta\left( c\right) -c\otimes 1-1\otimes c$ of $C^{+}\otimes C^{+}$ (proving that this element indeed lies in $C^{+}\otimes C^{+}$ is a nice exercise) by $\overline{\Delta}\left( c\right) $.


**§5.** Let $C\_{\left[ 1\right] }$, $C\_{\left[ 2\right] }$, $C\_{\left[ 3\right] }$, $...$ be graded coalgebras (I use the square brackets in $C\_{\left[ n\right] }$ to distinguish it from an $n$-th graded component of something). Then the direct sum $C\_{\left[ 1\right] }\oplus C\_{\left[ 2\right] }\oplus C\_{\left[ 3\right] }\oplus...$ is a graded coalgebra as well, but not connected even if $C\_{\left[ 1\right] }$, $C\_{\left[ 2\right] }$, $C\_{\left[ 3\right] }$, $...$ are. Instead, we can define a "connected graded sum" of connected graded coalgebras $C\_{\left[ 1\right] } $, $C\_{\left[ 2\right] }$, $C\_{\left[ 3\right] }$, $...$ as follows:


Consider the direct sum $k\oplus C\_{\left[ 1\right] }^{+}\oplus C\_{\left[ 2\right] }^{+}\oplus C\_{\left[ 3\right] }^{+}\oplus...$ of $k$-vector spaces, graded just like every other direct sum of graded $k$-vector spaces (the $k$, $C\_{\left[ 1\right] }^{+}$, $C\_{\left[ 2\right] }^{+}$, $C\_{\left[ 3\right] }^{+}$, $...$ are *not* the graded components). For every $n\in\mathbb{N}$, we can consider the $k$-vector space $C\_{\left[ n\right] }=k\oplus C\_{\left[ n\right] }^{+}$ as a subspace of this direct sum (because both $k$ and $C\_{\left[ n\right] }^{+}$ are summands). We define the coalgebra structure on $k\oplus C\_{\left[ 1\right] }^{+}\oplus C\_{\left[ 2\right] }^{+}\oplus C\_{\left[ 3\right] }^{+}\oplus...$ in such a way that $C\_{\left[ n\right] }$ becomes a *subcoalgebra* for every $n\in\mathbb{N}$. (This is easily seen to be well-defined.) This new graded coalgebra is connected.


**§6.** If $C\_{\left[ 1\right] }$, $C\_{\left[ 2\right] }$, $C\_{\left[ 3\right] }$, $...$ are locally finite connected graded coalgebras, then $k\oplus C\_{\left[ 1\right] }^{+}\oplus C\_{\left[ 2\right] }^{+}\oplus C\_{\left[ 3\right] }^{+}\oplus...$ needs not be locally finite. However, $k\oplus C\_{\left[ 1\right] \left( 1\right) }^{+}\oplus C\_{\left[ 2\right] \left( 2\right) }^{+}\oplus C\_{\left[ 3\right] \left( 3\right) }^{+}\oplus...$ is a locally finite connected graded coalgebra.


**§7.** Now to the counterexample. Let $k=\mathbb{R}$. For every positive $n\in\mathbb{N}$, let $\left( C\_{\left[ n\right] },f\_{n}\right) $ be a pair of a locally finite connected graded $k$-coalgebra $C\_{\left[ n\right] }$ and a map $f\_{n}\in\overline{C\_{\left[ n\right] }^{\ast}}^{2}$ of rank $n-1$. (Such a pair exists because we can take the pair $\left( T,f\_{n}\right) $ of §2.)


Then, $k\oplus C\_{\left[ 1\right] \left( 1\right) }^{+}\oplus C\_{\left[ 2\right] \left( 2\right) }^{+}\oplus C\_{\left[ 3\right] \left( 3\right) }^{+}\oplus...$ is a locally finite connected graded coalgebra. Denote it by $C$. Let $f\in\overline{C^{\ast}}$ be the $k$-linear map which acts as $0$ on $k$ and acts as $f\_{n}$ on each $C\_{\left[ n\right] }^{+}$ (we don't have to distinguish between $C\_{\left[ n\right] }^{+}$ and $C\_{\left[ n\right] \left( n\right) }^{+}$ here, since we are not talking about the grading). This map $f$ maps every primitive element of $C$ to $0$ (in fact, every primitive element of $C$ is a sum of primitive elements of some $C\_{\left[ n\right] \left( n\right) }^{+}$'s, and each of them is mapped to $0$ by the respective $f\_{n}$ because $f\_{n}\in\overline{C\_{\left[ n\right] }^{\ast} }^{2}$). If our assertion was true, then this map $f$ would lie in $\overline{C^{\ast}}^{2}$, and thus have a rank. Denote by $N$ this rank. We could thus write the map $f$ in the form $f=\sum\limits\_{i=1}^{N}\lambda \_{i}g\_{i}h\_{i}$ with $\lambda\_{i}\in k$, $g\_{i}\in\overline{C^{\ast}}$ and $h\_{i}\in\overline{C^{\ast}}$. Now, for every positive $n\in\mathbb{N}$, we could restrict both sides of the equation $f=\sum\limits\_{i=1}^{N}\lambda \_{i}g\_{i}h\_{i}$ to the subcoalgebra $k\oplus C\_{\left[ n\right] \left( n\right) }^{+}$ of $C$ (this subcoalgebra is isomorphic to $C\_{\left[ n\right] }$, at least if we don't care about the grading), and thus conclude that $f\_{n}=\sum\limits\_{i=1}^{N}\lambda\_{i}\left( g\_{i}\mid\_{k\oplus C\_{\left[ n\right] \left( n\right) }^{+}}\right) \left( h\_{i} \mid\_{k\oplus C\_{\left[ n\right] \left( n\right) }^{+}}\right) $, so that $f\_{n}$ has rank $\leq N$. But this is nonsense since the rank of $f\_{n}$ is known to be $n-1$, and $n-1>N$ for $n$ sufficiently large. This contradiction proves the counterexample.


