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In Higher Algebra by Lurie, we define the absolute cotangent complex $L_{A}$ through the composition $C\stackrel{\triangle}{\longrightarrow} Fun(\triangle^{1},C)\stackrel{F}{\longrightarrow}T_{C}$ where $T_{C}$ is the tangent bundle to $C$，and the functor $F$ restricts to the fiber over $B$ as the infinite suspension functor $\sum^{\infty}: C_{/B}\rightarrow Sp(C_{/B})$. And the absolute cotangent complex $L_{A}$ is exactly the infinite suspension of $B\stackrel{id}{\longrightarrow} B\in C_{/B}$ because the diagonal functor sends $B\in C$ to identity map in $Fun(\triangle^{1},C)$.

And so my $\textbf{question}$ is:

$\textbf{Is the relative cotangent complex $L_{B/A}$ the infinite suspension of $A\rightarrow B\in C_{/B}$?}$

Here is my attempt on this question:

The relative cotangent complex $L_{B/A}$ is identified with the absolute cotangent complex in $B\in C_{A/}$ by corollary 7.3.3.15 in HA, so we can see that the relative cotangent complex $L_{B/A}$ is the infinite suspension of $(A\rightarrow B)\stackrel{id}{\longrightarrow} (A\rightarrow B)\in (C_{A/})_{/B}$, so I need to show that such inifinite suspension coincides with the suspension of $A\rightarrow B\in C_{/B}$.

In short, it suffices to prove the existence of the following commutative diagram:

enter image description here

It also suffices to prove the existence of the following diagram:

enter image description here

,which seems to be some properties of infinite suspension functor with respect to under-category.

Also I noticed that the existence of the following commutative diagram, which is kind of the dual version of what I want to get: enter image description here

But I don't know whether there exists a mechanism to have the commutative diagram of the adjoint functor?

Any thoughts on this are very welcome!

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It is not, for example note that if this were true, the case $A=B$ would give you that $L_{B/B}$ is the infinite suspension of $\mathrm{id}: B\to B$. But that's the absolute cotangent complex $L_B$, whereas $L_{B/B} =0$. You really have to take the absolute cotangent complex in the slice under $A$.

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  • $\begingroup$ That’ a great example, do you have an idea on the infinite suspension of $f:A\rightarrow B$ in $Sp(C_{B})$ is? I guess it’s still related with the cotangente complex, I should probably try $f_{*}L_{A}$? $\endgroup$
    – Yang
    Commented Aug 30 at 7:33
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    $\begingroup$ I think the infinite suspension of $A\to B$ should agree with the base-change $f^* L_A$, yes. It agrees with the reduced suspension $\Sigma^\infty$ (I would write the other one $\Sigma^\infty_+$) of the pointed object $A\amalg B \to B$ in $\mathcal{C}_{/B}$, and I seem to recall that the reduced suspension of any pointed $B\to C \xrightarrow{g} B$ agrees with the "cotangent fiber" $g^* L_{C/B}$. For $C=A\amalg B$ we should have $g^* L_{A\amalg B/B} = f^* L_A$. $\endgroup$
    – Achim Krause
    Commented Aug 30 at 8:51
  • $\begingroup$ Thanks a lot, would you please share some references on the claim 'the reduced suspension of the pointed object $B\rightarrow C\stackrel{g}{\longrightarrow} B$ agrees with the 'cotangent fiber', I tried to figure out what's going on here but failed, it confuses me a little because $L_{C.B}$ should lie in $Sp(B)$, and its pullback along g seems to take it to somewhere else. $\endgroup$
    – Yang
    Commented Aug 31 at 14:24
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    $\begingroup$ Sorry, I should have written pushforward. $L_{C/B}$ should lie in $\mathrm{Sp}(C)$, and $g_*$ takes it into $\mathrm{Sp}(B)$. I'm afraid I don't have a reference at the moment, especially not in the generality you're thinking about (all my intuition for this comes from rings and modules). But I would be surprised if this can't be extracted from the corresponding section of Higher Algebra... $\endgroup$
    – Achim Krause
    Commented Aug 31 at 14:41
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    $\begingroup$ Yes! So if $B$ is a ring and $C$ a $B$-algebra, it would be a $C$-module, and $g_* L_{C/B}$ would be $B\otimes_C L_{C/B}$. $\endgroup$
    – Achim Krause
    Commented Aug 31 at 15:28

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