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LSpice
LSpice
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Is every finite spectrum $X$ $K(h)$-locally equivalent to a finite spectrum $Y$ with $dim$\dim (K(h)_\ast Y) = dim\dim ((H\mathbb F_p)_\ast Y)$?

Let $X$ be a finite spectrum and $K = K(h)$ be the $h$th Morava $K$-theory at the prime $p$. Then $dim_{K_\ast} K_\ast X$$\dim_{K_\ast} K_\ast X$ is increasing in $h$increasing in $h$, and eventually constant at $dim H_\ast(X,\mathbb F_p)$$\dim H_\ast(X,\mathbb F_p)$.

Question: Let $X$ be a finite spectrum. Does there exists a finite spectrum $Y$ which is $K(h)$-locally equivalent to $X$ such that $dim_{K_\ast}K_\ast Y = dim_{\mathbb F_p} H_\ast(Y;\mathbb F_p)$$\dim_{K_\ast}K_\ast Y = \dim_{\mathbb F_p} H_\ast(Y;\mathbb F_p)$?

I think the answer is probably no. One way this could be so comes when $dim_{K_\ast} K_\ast X = 1$$\dim_{K_\ast} K_\ast X = 1$ : if there is a nontrivial element of $Pic(Sp_{K(h)})$$\operatorname{Pic}(\operatorname{Sp}_{K(h)})$ which is the localization of a finite spectrum, then the answer is no. But I'm not sure if such elements exist.

Is every finite spectrum $X$ $K(h)$-locally equivalent to a finite spectrum $Y$ with $dim (K(h)_\ast Y) = dim ((H\mathbb F_p)_\ast Y)$?

Let $X$ be a finite spectrum and $K = K(h)$ be the $h$th Morava $K$-theory at the prime $p$. Then $dim_{K_\ast} K_\ast X$ is increasing in $h$, and eventually constant at $dim H_\ast(X,\mathbb F_p)$.

Question: Let $X$ be a finite spectrum. Does there exists a finite spectrum $Y$ which is $K(h)$-locally equivalent to $X$ such that $dim_{K_\ast}K_\ast Y = dim_{\mathbb F_p} H_\ast(Y;\mathbb F_p)$?

I think the answer is probably no. One way this could be so comes when $dim_{K_\ast} K_\ast X = 1$ : if there is a nontrivial element of $Pic(Sp_{K(h)})$ which is the localization of a finite spectrum, then the answer is no. But I'm not sure if such elements exist.

Is every finite spectrum $X$ $K(h)$-locally equivalent to a finite spectrum $Y$ with $\dim (K(h)_\ast Y) = \dim ((H\mathbb F_p)_\ast Y)$?

Let $X$ be a finite spectrum and $K = K(h)$ be the $h$th Morava $K$-theory at the prime $p$. Then $\dim_{K_\ast} K_\ast X$ is increasing in $h$, and eventually constant at $\dim H_\ast(X,\mathbb F_p)$.

Question: Let $X$ be a finite spectrum. Does there exists a finite spectrum $Y$ which is $K(h)$-locally equivalent to $X$ such that $\dim_{K_\ast}K_\ast Y = \dim_{\mathbb F_p} H_\ast(Y;\mathbb F_p)$?

I think the answer is probably no. One way this could be so comes when $\dim_{K_\ast} K_\ast X = 1$ : if there is a nontrivial element of $\operatorname{Pic}(\operatorname{Sp}_{K(h)})$ which is the localization of a finite spectrum, then the answer is no. But I'm not sure if such elements exist.

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Tim Campion
Tim Campion
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Is every finite spectrum $X$ $K(h)$-locally equivalent to a finite spectrum $Y$ with $dim (K(h)_\ast Y) = dim ((H\mathbb F_p)_\ast Y)$?

Let $X$ be a finite spectrum and $K = K(h)$ be the $h$th Morava $K$-theory at the prime $p$. Then $dim_{K_\ast} K_\ast X$ is increasing in $h$, and eventually constant at $dim H_\ast(X,\mathbb F_p)$.

Question: Let $X$ be a finite spectrum. Does there exists a finite spectrum $Y$ which is $K(h)$-locally equivalent to $X$ such that $dim_{K_\ast}K_\ast Y = dim_{\mathbb F_p} H_\ast(Y;\mathbb F_p)$?

I think the answer is probably no. One way this could be so comes when $dim_{K_\ast} K_\ast X = 1$ : if there is a nontrivial element of $Pic(Sp_{K(h)})$ which is the localization of a finite spectrum, then the answer is no. But I'm not sure if such elements exist.