if a symmetric spectrum X (semistable if it helps) is such that $\pi_n(X)$ is rational, can I represent any element of $\pi_n(X)$ by a map $\Sigma^n H\mathbb{Q} \to X$?

@BrunoStonek correct me if I'm wrong, but if X is rational, then X = L_Q X; so any element of pi_n(X), represented by a map Σ^n S --> X, factors through L_Q Σ^n S --> X, which is a map Σ^n HQ --> X

@skd but I'm not assuming that $X$ is rational, only that $\pi_n(X)$ (for a fixed $n$) is rational

ah

well, disregarding point-set issues if that's ok

take an element alpha in pi_n(X)

then there's a triangular commutative diagram S^n -2-> S^n -(alpha/2)-> X in which the composite is alpha

since you can divide alpha by arbitrary powers of 2, you can extend this to a map from the filtered system S^n -2-> S^n -2-> S^n -2-> .... -> X

I don't know why I'm not using tex

so you get a map from the filtered colimit, which is $\Sigma^n \mathbb{S}[\frac 1 2]$, to $X$

repeat for all primes, and you get a map from $\Sigma^n (\mathbb{S}[\frac 1 2, \frac 1 3, \cdots])$ to $X$, and the former is of course $\Sigma^n H\mathbb{Q}$.

however, I think the neighbouring homotopy groups will affect the uniqueness of this map

oh, maybe not

thanks. I didn't think of seeing HQ as the sphere spectrum with primes inverted

Is there a characterization of when a recollement is presentable? I'm happy to assume both strata are presentable, but I do not want to assume that the gluing functor is accessible (though that is sufficient). For example, how could we show that Sp^C_2 is accessible if we just knew it as the recollement of Sp and local systems on BC_2 using the Tate construction to glue?

(maybe @SaulGlasman has thought about this?)

Suppose a person is interested in algebraic geometry. Certainly, sometimes it is useful to enlarge the category of schemes and work in the category of derived schemes (for instance, to obtain correct fiber products), so derived algebraic geometry is useful for understanding algebraic geometry. Is the same true for spectral algebraic geometry? is there any reason for an algebraic geometer to consider the category of spectral schemes?

Is every Bousfield localization of spectra a localization with respect to a ring spectrum? I.e., given a spectrum X, is there a ring spectrum with the same bousfield class?

doesn't End(X) have the same Bousfield class as X?

Hmm.. this question seems to say the answer is no: mathoverflow.net/questions/151968/…

(it is talking about HF_p-local spectra rather than spectra but that shouldn't matter)

by "Bousfield class", does the asker mean Bousfield class of a ring, or of an arbitrary spectrum?

The Bousfield class of a spectrum $X$ is the class of spectra $Y$ such that $X\wedge Y\not\approx 0$.

oh, I thought the Bousfield class of X was the class of X' for which X ^ Y = 0 if and only if X' ^ Y = 0

Lol maybe. In either case, "$X$ and $X'$ have the same Bousfield class" means the same thing under either definition.

Has this happened to anyone? I compiled and latex inserted some huge amount of vertical space before and after a display-math thing for no apparent reason... Help?

Even weirder: the spacing only occurs after I type 4 lines of text after the display math... but not 3 lines of text. Wtf...

In computation how do we play the isotropy separation sequence? For understanding a G spectrum X, I need to know about its geometric fixed point, which is usually doable. But the first term EP_+ \wedge X, should be determined by X as a H-spectrum for subgroups H, but explicitly how?

Like how should I compute \pi_*^G of that guy if I know X as H-spectrum very well