# matrix ring spectra

I am trying to understand matrix ring spectra. Apparently, I have two different definitions of those and I did not manage to show that they are equivalent - maybe they even are not in the general case.

Given a connective (commutative) symmetric ring spectrum $R$, we define the $k$-th space of $M_n(R)$ to be the following space of pointed maps $$[M_n(R)]_k = map_*(n_+, n_+ \wedge R_k)$$ where the multiplication is given by $$n_+ \to n_+ \wedge R_l \to n_+ \wedge R_k \wedge R_l \to n_+ \wedge R_{k+l}$$ i.e. apply the first map, apply the second map wedge identity, then multiply in $R$. The definition is used in this paper to define the algebraic $K$-theory of $R$ by considering $GL_1(M_n(R))$. The latter has the right homotopy type if $R$ is convergent (see the reference). Correct me if I am wrong, but this should be stably equivalent to $End_R(n_+ \wedge R)$, where we consider $n_+ \wedge R$ as an $R$-module spectrum in the obvious way.

My second definition works as follows: Take $R \times \dots \times R$ as an $R$-module with the "diagonal" $R$-action and define $M_n'(R) = End_R(R \times \dots \times R)$. At least if $R$ is semistable, the canonical map $n_+ \wedge R \to R \times \dots \times R$ should be a stable equivalence, but according to the answer to my last question this is not enough to deduce that the endomorphism spectra are stably equivalent.

Does the second definition make sense? Would it be enough to demand that $R$ is fibrant-cofibrant to get the equivalence? Is there a more general condition to achieve equivalence in this situation?

The way I see those two versions is that the space $M_n(R)$ contains matrices with just one non-vanishing column (or row?), whereas $M_n'(R)$ contains full matrices. Going over to homotopy classes of maps from a space into $M_n(R)$ adds the "elementary matrices" together (consider for example a loop running through two different copies of $R$). Thus, up to homotopy it should look like $M_n'(R)$.

• I did not understand your first definition of matrix endomorphism but for the second one I would say it is better if to modify it by taking the wedge RvRv...vR instead of RxR...xR, in suitable model category of spectra RvRv...vR is fibrant-cofibrant. – Ilias A. Jul 4 '14 at 21:32
• @Fedotov: That would coincide with the first definition. $n_+$ is the set of numbers $\{1, \dots, n\}$ with an added basepoint. – Ulrich Pennig Jul 5 '14 at 6:42
• Perfect, since RvR...vR and RxR....xR are stably equivalent, the first definition is more adequate because RvR.....vR is cofibrant fibrant as R-Module (in a good choice of model category) that is probably why the authors of your cited article have made this choice therefore End_R(RvR..vR) is a homotopy invariant object. The cofibrancy property is not well behaved with the cartesian product in the model category of spectra. – Ilias A. Jul 5 '14 at 7:06
• @Fedotov: The problem is that I have something that I want to understand and that maps nicely into the second definition, but not into the first. From what you say, I guess, I have to figure out if it also maps nicely into some cofibrant replacement of $R \times \dots \times R$, then. – Ulrich Pennig Jul 5 '14 at 7:20
• RvR....vR is a candidate for the cofibrant replacement of RxR....xR. – Ilias A. Jul 5 '14 at 7:22