# coproduct of the homology of iterated loop space on spheres

Let $\Omega^{n+1}S^{n+1}$ be the base-pointed $(n+1)$-iterated loop space on the $(n+1)$-sphere. In the paper The homology of $\mathcal{C}_{n+1}$-spaces, $n\geq 0$, F. Cohen, Lecture notes in mathematics Vol. 533, page 226, Theorem 3.2, it is proved that as a $AR_n\Lambda_n$-Hopf algebra, $$H_*(\Omega^{n+1}S^{n+1};\mathbb{Z}_p)\cong GW_n H_*(S^0;\mathbb{Z}_p)$$ for primes $p\geq 2$.

Question: In page 223, it is said that "the coproduct are determined by the diagonal Cartan formulas (page 213, (4))". I do not understand how to write the coproduct of $H_*(\Omega^{n+1}S^{n+1};\mathbb{Z}_p)$ explicitly?

In the paper On the homology of configuration spaces, Section 4, as an associative and commutative $\mathbb{Z}_p$-algebra, a generator set of $H_*(\Omega^{n+1}S^{n+1};\mathbb{Z}_p)$ is given. I want to know how to obtain the coproduct formulas of $H_*(\Omega^{n+1}S^{n+1};\mathbb{Z}_p)$ as a coalgebra.

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For the answer below, I have a question.

In the paper On the homology of configuration spaces, Section 4.1, a basis for the graded vector space $H_*(\Omega^m S^m;\mathbb{Z}_2)$ is given as

$(*)$ $$u_0, u_1, Q_I u_0, Q_I u_1$$ where $0,1$ denote the two generators of $H_0(S^0;\mathbb{Z}_2)$ and $Q_I=Q_{i_1}Q_{i_2}\cdots Q_{i_r}$ for $I=(i_1,i_2,\cdots, i_r)$, $0<i_1\leq i_2\leq \cdots\leq i_r<m$.

By the Cartan formula (cf. The homology of $\mathcal{C}_{n+1}$-spaces, $n\geq 0$, F. Cohen, page 213)

In the answer below, we calculated that

$(**)$ $$\Delta_* Q_I u_0=\sum_{t_1=0}^{i_1}\sum_{t_2=0}^{i_2}\cdots \sum _{t_r=0}^{i_r}Q_{t_1}Q_{t_2}\cdots Q_{t_r}u_0\otimes Q_{i_1-t_1}Q_{i_2-t_2}\cdots Q_{i_r-t_r}u_0,$$

$(***)$ $$\Delta_* Q_I u_1=\sum_{t_1=0}^{i_1}\sum_{t_2=0}^{i_2}\cdots \sum _{t_r=0}^{i_r}Q_{t_1}Q_{t_2}\cdots Q_{t_r}u_1\otimes Q_{i_1-t_1}Q_{i_2-t_2}\cdots Q_{i_r-t_r}u_1.$$

Is this result correct or wrong?

Question: The problem is that $Q_{t_1}Q_{t_2}\cdots Q_{t_r}u_0$, $Q_{i_1-t_1}Q_{i_2-t_2}\cdots Q_{i_r-t_r}u_0$, $Q_{t_1}Q_{t_2}\cdots Q_{t_r}u_1$, $Q_{i_1-t_1}Q_{i_2-t_2}\cdots Q_{i_r-t_r}u_1$ are not included in the basis. It does not make sense. Why?

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I find the Adem relations for $p=2$

i.e.

for $t_1>2t_2$,

$$Q_{t_1}Q_{t_2}=\sum_{t_1/2 \leq i\leq t_1-t_2-1}(-1)^{t_1+i}(2i-t_1,t_1-t_2-i-1)Q^{t_1+t_2-i}Q^i.$$

How to apply the Adem relation to simplify equations $(**)$, $(***)$ to sums of tensor products of basis given in $(*)$?

• Commented on an identical question: you have to use the Adem relations. – Peter May Aug 29 '15 at 2:04

The case for odd primes being slightly more complicated to write down due to the Bockstein, we deal with the case $p=2$ here, although the odd prime cases work essentially in the same way. Since it is a Hopf algebra, we have $$\Delta (x\cdot y) =\Delta (x)\cdot \Delta (y).$$ Thus it suffices to know the coproduct on algebra generators. However, the algebra generators are of the form $Q^{I_1}Q^{I_2}\cdots Q^{I_r}([1])$, and $[-1]$. The coproduct on $[\pm1]$ is given by $\Delta ([\pm 1])=[\pm 1]\otimes [\pm 1]$, and the repeated use of Cartan formula gives $$\Delta (Q^{I_1}Q^{I_2}\cdots Q^{I_r}([1]))=\Sigma _{J_1,J_2,\ldots J_r} Q^{J_1}Q^{J_2}\cdots Q^{J_r}([1])\otimes Q^{I_1-J_1}Q^{I_2-J_2}\cdots Q^{I_r-J_r}([1]).$$
• Thanks! Does $[-1]$ mean the base point of $S^0$? Why the $Q^I$'s do not act both on $[1]$ and $[-1]$, but only act on $[1]$? – QSR Aug 10 '15 at 3:13
• Are the algebra generators of the form $Q^{I_1}Q^{I_2}\cdots Q^{I_r}([1])$ and $Q^{I_1}Q^{I_2}\cdots Q^{I_r}([-1])$, instead of $Q^{I_1}Q^{I_2}\cdots Q^{I_r}([1])$ and $[-1]$? – QSR Aug 10 '15 at 3:16
• See www.math.uchicago.edu/~may/BOOKS/homo_iter.pdf chapter 4. The base point of $S^0$ is [0]. [-1] is the component of the degree -1 map. – user43326 Aug 10 '15 at 5:38