Recall that a non-unital ring $R$ is called Tor-unital if $Tor^1_{R_+}(\mathbb Z,\mathbb Z) \cong 0$ where $R_+$ is the unitalization of $R$. See e.g. https://arxiv.org/pdf/1610.04998.pdf. If $R$ is Tor-unital, does $\bigsqcup_n BGL_n(R)$ have a natral $E_\infty$-algebra structure?
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2$\begingroup$ I'm not familiar with Tor-unital rings, but the E_∞-structure of $\coprod_n BGL_n(R)$ is secretly coming from the tensor product of $R$-modules. Do you know if modules over a (commutative?) Tor-unital ring have a tensor product? And how are you defining $GL_n(R)$ for a non-unital ring? $\endgroup$– Denis NardinCommented Jan 5, 2019 at 6:56
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$\begingroup$ Don’t we usually use the direct sum, not the tensor product? $\endgroup$– Dylan WilsonCommented Jan 5, 2019 at 13:51
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$\begingroup$ @DylanWilson I guess I interpreted "E_∞-algebra" as "E_∞-ring" structure (mainly because the direct sum structure is always there for trivial reasons), but you're right, the OP might just mean that. Although in that case I don't understand why the Tor-unitality is relevant. $\endgroup$– Denis NardinCommented Jan 5, 2019 at 14:10
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$\begingroup$ If the ring is not unital, the category of $R$ modules with direct sum might not be symmetric monoidal so $\sqcup BGL_n(R)$ might not be an $E_\infty$-algebra in spaces. $GL_n(R)$ of a non-unital ring can be thought of as a congruence subgroup. In particular, it might not contain the symmetric group on $n$ letters which is needed to prove that the category of $R$-modules to be symmetric monoidal. Possibly Tor unitality will imply that in some derived sense, $GL_n(R)$ contains the symmetric group. Tor unitality is equivalent to excision in algebraic $K$-theory. $\endgroup$– eeeeeeCommented Jan 5, 2019 at 14:58
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$\begingroup$ One way to define $GL_n(R)$ for a non-unital ring is to realize $R$ as an ideal in a unital ring $K$ and let $GL_n(R)$ be the kernel of the map $GL_n(K) \to GL_n(K/R)$. This is independent of the ring $K$. $K$ exists as there is a unital ring structure on $R \oplus \mathbb Z$ for any non-unital ring $R$. $\endgroup$– eeeeeeCommented Jan 6, 2019 at 10:32
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