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Let $X$ be a CW complex and let $\Sigma^\infty X$ denote its suspension spectrum. By definition, the $n$th singular homology group of $\Sigma^\infty X$ with coefficients in $\mathbb{Z}$ is $\pi_n(\Sigma^\infty X \wedge H\mathbb{Z})$.

Now, connective spectra are in bijection with infinite loop spaces. The infinite loop space corresponding to $\Sigma^\infty X$ is $QX := \Omega^\infty \Sigma^ \infty X = \lim_{n \to \infty} \Omega^n \Sigma^n X$. Is the homology of $\Sigma^\infty X$ (as a spectrum) isomorphic to the homology of $QX$ (as a CW complex)? I would expect this.

But what confuses me is a theorem is Rudyak's On Thom spectra, orientability, and cobordism (Theorem 7.11.(i)). Rudyak states that for every rational spectrum $E_\mathbb{Q}$ the Hurewicz map $\pi_\ast(E_\mathbb{Q}) \rightarrow H_\ast(E_\mathbb{Q})$ is an isomorphism. This seems to prove that the homotopy/homology groups of spectra do not agree with homotopy/homology groups of their corresponding infinite loop spaces.

First, $\pi_\ast(Q X) \otimes \mathbb{Q}$ is isomorphic to the stable rational homotopy $\pi_\ast^S(X) \otimes \mathbb{Q}$ of $X$. Then, by the Milnor-Moore theorem, $H_\ast(QX,\mathbb{Q}) \cong \mathbb{Q}[\pi_\ast^S(X) \otimes \mathbb{Q}]$. So we'd have that $H_\ast(X,\mathbb{Q})$ is the free commutative-graded algebra on the rational stable homotopy groups of $X$ for any CW complex $X$. But it is easy to write down examples of CW complexes whose rational homology is not free e.g. projective spaces. So what, then, is the relationship between $\pi_\ast(\Sigma^\infty X \wedge H\mathbb{Z})$ and $H_\ast(QX, \mathbb{Z})$?

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    $\begingroup$ the homology of the infinite loop space is the same thing as the homology of its suspension spectrum, something which can be directly worked out using definitions. this is larger than the homology of the associated omega-spectrum, since the latter only comes from those homotopy classes of maps (at level zero) which deloop infinitely many times, i.e. the ones which can be refined to maps of E-infinity algebras. $\endgroup$ – xir Apr 5 '19 at 19:19
  • $\begingroup$ Sorry could you explain how exactly do you apply Milnor - Moore? $\endgroup$ – მამუკა ჯიბლაძე Apr 5 '19 at 20:09
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    $\begingroup$ Sorry. The Milnor-Moore theorem says that the rational homology of an associative and unital H-space $X$ is the universal enveloping algebra of $\pi_\ast(X) \otimes \mathbb{Q}$, where $\pi_\ast(X) \otimes \mathbb{Q}$ has the structure of a graded Lie algebra via Whitehead products. $\endgroup$ – user137162 Apr 5 '19 at 21:32
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$\newcommand{\H}{\mathrm{H}} \newcommand{\Z}{\mathbf{Z}}$Let $X$ be a space. Then the $E$-(co)homology of $X$ is the same as the $E$-(co)homology of its suspension spectrum, i.e., $E_\ast(X) \cong E_\ast(\Sigma^\infty_+ X)$ (and you remove the basepoint in $\Sigma^\infty_+ X$ to get reduced $E$-homology). In the case when $E = \H\Z$, this says that $\H_\ast(X;\Z) \cong \H_\ast(\Sigma^\infty_+ X;\Z)$. In the case when $E$ is the sphere spectrum, this says that the stable homotopy groups of $X$, i.e., the unstable homotopy groups of $QX$, are the same as the homotopy groups of the spectrum $\Sigma^\infty X$.

Let's now turn to the homology of $QX$. We know that $\H_\ast(QX;\Z) \cong \H_\ast(\Sigma^\infty_+ QX;\Z)$, so you need to understand $\Sigma^\infty_+ QX$. A theorem of Snaith's says that there is an equivalence $$\Sigma^\infty_+ QX \simeq \bigvee_{n\geq 0} (\Sigma^\infty X)^{\wedge n}_{h\Sigma_n},$$ so we find that $$\H_\ast(QX;\Z)\cong \bigvee_{n\geq 0} \H_\ast((\Sigma^\infty X)^{\wedge n}_{h\Sigma_n};\Z).$$ The groups $\H_\ast((\Sigma^\infty X)^{\wedge n}_{h\Sigma_n};\Z)$ can be computed by a homotopy orbits spectral sequence: $$E_2^{s,t} = \H_\ast(\Sigma_n; \H_\ast((\Sigma^\infty X)^{\wedge n};\Z)) \Rightarrow \H_\ast((\Sigma^\infty X)^{\wedge n}_{h\Sigma_n};\Z).$$ If $\Z$ is replaced by $\mathbf{Q}$, then this spectral sequence degenerates (the $E_2$-page vanishes for $s>0$), and you find that $\H_\ast((\Sigma^\infty X)^{\wedge n}_{h\Sigma_n};\mathbf{Q}) \cong \H_\ast(\Sigma^\infty X;\mathbf{Q})^{\otimes n}_{\Sigma_n}$, i.e., $\H_\ast(QX;\mathbf{Q}) \cong \mathrm{Sym}^\ast \H_\ast(\Sigma^\infty X;\mathbf{Q})$. By the way, something special happens with $\mathbf{Q}$-cohomology of spectra: since the $\mathbf{Q}$-localization of the sphere spectrum is $\H\mathbf{Q}$, you find that $\pi_\ast(F_\mathbf{Q}) \cong \H_\ast(F_\mathbf{Q};\mathbf{Q}) \cong \H_\ast(F;\mathbf{Q})$ for any spectrum. (This is more generally true for any smashing localization.) (Edit: Whoops, I didn't actually finish my answer.) This means that $\H_\ast(\Sigma^\infty X;\mathbf{Q}) \cong \pi_\ast(\Sigma^\infty X_\mathbf{Q}) \cong \pi_\ast(\Sigma^\infty X) \otimes \mathbf{Q} = \pi_\ast^s(X) \otimes \mathbf{Q}$, so you find that $$\H_\ast(QX;\mathbf{Q}) \cong \mathrm{Sym}^\ast(\pi_\ast^s(X) \otimes \mathbf{Q}),$$ as you said in your post. The homology of $QX$ with coefficients in a general homology theory $E$ is a lot more complicated: if $E$ is a structured ring spectrum, it describes what are known as Dyer-Lashof operations for $E$.

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Let us denote by $B^\infty : \infty\rm{LoopSpaces} \stackrel{\sim}{\to} \rm{ConnectiveSpectra}$ the equivalence of ($\infty$-)categories you hinted in your question. Let us first consider the case of reduced homology of a pointed CW-complex $X$, as formulas are slightly cleaner. Unwinding the definitions, we see that

$$\pi_n(\Sigma^\infty X \wedge H\mathbb Z) = \pi_n(B^\infty (QX) \wedge H\mathbb Z)$$ and $$\widetilde H_*(QX;\mathbb Z) = \pi_n(\Sigma^\infty (QX) \wedge H\mathbb Z) = \pi_n(B^\infty(QQX) \wedge H\mathbb Z)$$

$$H_*(QX;\mathbb Z) = \pi_n(\Sigma^\infty_+ (QX) \wedge H\mathbb Z) = \pi_n(B^\infty(Q((QX)_+)) \wedge H\mathbb Z).$$

So, essentially the difference lies in how many times you apply the $Q$ functor to $X$ before taking the homology of the associated connective spectrum.

When $X$ is unpointed, the fomulas above become $$\pi_n(\Sigma^\infty_+ X \wedge H\mathbb Z) = \pi_n(B^\infty (Q(X_+)) \wedge H\mathbb Z)$$ and $$\widetilde H_*(Q(X_+);\mathbb Z) = \pi_n(\Sigma^\infty (Q(X_+)) \wedge H\mathbb Z) = \pi_n(B^\infty(QQ(X_+)) \wedge H\mathbb Z)$$

$$H_*(Q(X_+);\mathbb Z) = \pi_n(\Sigma^\infty_+ (Q(X_+)) \wedge H\mathbb Z) = \pi_n(B^\infty(Q(Q(X_+)_+)) \wedge H\mathbb Z).$$

One way to clarify what happens is to keep track of which structure you want to consider on $QX$. By definition, the functor $Q$ carries pointed spaces to infinite loop spaces (that is, spaces with some extra structure). One can say that computing the homology of $X$, is equivalent to compute the homology of $QX$ as an infinite loop space (that is, the homology of the spectrum $B^\infty Q X$), whereas if you forget this extra structure on $QX$ and just look at its underlying pointed space, to know its homology you'll have to apply again the $Q$ functor (in order to land again in infinite loop spaces), and then compute its homology in the appropriate ($\infty$-)category.

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First of all, I think the concern of the OP is the fact that the homology of the infinite loop space doesn't agree with that of the spectra. So let's start with this. Let $Y$ be a spectrum, then it is homology is related to its infinite loop spaces' homology by $$H_*(Y)=\lim _i H_{*+i}(\underline{Y}_{i+j})$$ where $Y_k$ is the $k$-th infinite loop space associated to $Y$. Notably we have $Y_{k-1}=\Omega Y_k$, so it should be more or less clear that the homology of a spectrum can almost never agree with that of the associated infinite loop spaces.

Now, let's get to $H_*(QX)$. As has been noted, its rational homology is easy. So basically we need to know $p$-complete homology for all prime numbers $p$ to recover integral homology. In other words, we need to know the Bockstein spectral sequence for all prime $p$. The input of the Bockstein spectral sequence $H_*(QX;Z/p)$ was computed by Kudo-Araki for $p=2$, Dyer-Lashof for odd prime. The rest was worked out by May in Chapter 3, Theorem 3.12 of Frederick R. Cohen, Thomas J. Lada, J. Peter May "The homology of iterated loop space" Springer Lecture Notes in Mathematics 533.

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