I was wondering if there exists a model structure on the category of nonnegative differential graded algebras with homological grading. To be more precise: Let $Ch_{k}$ the model category of nonnegative chain complexes over a field $k$ (any characteristic). An object is of the form $$M_{0}\leftarrow M_{1}\leftarrow \dots$$ Weak equivalences are quasiisomorphisms and fibrations are morphisms of chain complexes $f_{\bullet}:M_{\bullet}\rightarrow N_{\bullet} $ such that $f_{\bullet}$ is surjective for $\bullet>0$. Is there a model structure on the category of differential graded algebras, and more generally is there a model structure on the category of $P$algebras for some cofibrant operad $P$? References are welcome. Thank you very much.

1$\begingroup$ Nonnegative chain complexes over an arbitrary ground commutative ring form a symmetric monoidal model category satisfying the monoid axiom with the structure you indicate, hence the first part of your question has a positive answer by SchwedeShipley. As for your second question, I think the answer may be positive too, I know it is for nonsymmetric operads at least, see Spitzweck's old preprint for the general case. $\endgroup$ – Fernando Muro May 21 '15 at 8:07

$\begingroup$ To add to Fernando's answer, have a look at BergerMoerdijk Axiomatic Homotopy for Operads. In this case I'd look for a path object rather than trying to follow Spitzweck and do the filtration style argument for a general cofibrant operad $P$. Those filtrations can get very hairy and often you only end up with a semimodel structure not a full model structure. $\endgroup$ – David White May 21 '15 at 19:11

$\begingroup$ Perhaps late, but have you seen this paper by V. Hinich? $\endgroup$ – Pedro Tamaroff Mar 27 '18 at 0:06

$\begingroup$ And this paper by Jardine, for DGalgebras? $\endgroup$ – Pedro Tamaroff Mar 27 '18 at 1:02
Yes, this is possible.
Because of the way you worded your question, I'm going to assume you already know the existence of the model structure for unbounded $P$algebras and use that in the proof. I also want to assume that your operad is concentrated in nonnegative degrees so that the free bounded algebra functor behaves as we expect it to.
Theorem 3.3 of Crans, Quillen closed model structures for sheaves gives criteria for transfer of a cofibrantly generated model category structure along an adjoint that are easily checkable in this case.
Let $C$ be a cofibrantly generated model category with $I$ (respectively $J$) the set of generating (trivial) cofibrations and let $R:D\to C$ be a right adjoint with left adjoint $L$. You want a model category on $D$ where the fibrations and weak equivalences are created by $R$.
The conditions you have to verify are that
 $D$ has small limits and colimits,
 a smallness criterion on $L(I)$ and $L(J)$, and
 relative $L(J)$cell complexes become $C$ weak equivalences under $R$.
In this case $C$ is bounded chains, $D$ is bounded $P$algebras, $R$ is the forgetful functor, and $L$ is the free algebra functor. We will exploit the fact that we already know something about unbounded algebras to show that we have condition (3.)
You presumably know (1.) or you wouldn't even be trying. (2.) is easy in algebraic categories and I'll skip it. You basically already know it anyway from the existence of the transferred model structure in the unbounded case.
For (3.), note that the inclusion $i$ of bounded algebras into unbounded algebras reflects weak equivalences: $R_{\text{unbounded}}i(f)$ is a weak equivalence implies $R_{\text{bounded}}(f)$ is a weak equivalence. So it suffices to check that $i$ applied to an $L_\text{bounded}(J_\text{bounded})$relative cell complex is a weak equivalence under $R_\text{unbounded}$.
The inclusion $i$ is a left adjoint and so preserves colimits, so such a morphism is an $iL_\text{bounded}(J_\text{bounded})$relative cell complex. But $iL_\text{bounded}$ is the restriction of $L_\text{unbounded}$ and $J_\text{bounded}\subset J_\text{unbounded}$ so any such morphism is an $L_\text{unbounded}(J_\text{unbounded})$relative cell complex, and thus a trivial cofibration in the model structure on unbounded algebras, and thus a weak equivalence under $R_\text{unbounded}$.

$\begingroup$ $iL_{\text{bounded}}$ is not the restriction of $L_{\text{unbounded}}$. In the bounded case, you don't ask fibrations to be surjective in degree $0$. Actually, it wouldn't work if you did. The inclusion is a left Quillen functor, not a right Quillen functor (although it is a right adjoint). The model structure on bounded complexes is not transferred from unbounded complexes. $\endgroup$ – Fernando Muro May 22 '15 at 8:31

1$\begingroup$ @FernandoMuro the statement you take issue with is categorical and has no modelcategorical content. $L$ is the free algebra functor. Saying $iL_\text{bounded}$ is the restriction of $L_\text{unbounded}$ just means that the free unbounded $P$algebra on a bounded chain complex $V$ coincides with the free bounded $P$algebra on $V$, considered as an unbounded $P$algebra. $\endgroup$ – Gabriel C. DrummondCole May 22 '15 at 9:06

1$\begingroup$ I also don't understand why you brought up transferring from unbounded to bounded complexes since I didn't do that in my answer. $\endgroup$ – Gabriel C. DrummondCole May 22 '15 at 11:35

$\begingroup$ Sorry, I think I misunderstood your notation. In any case, I don't think you can make it easier than in the comments since the bounded case isn't more complicated than the unbounded case, that you assume. $\endgroup$ – Fernando Muro May 22 '15 at 12:55

$\begingroup$ @FernandoMuro I added another little paragraph, is it clearer? About bounded/unbounded, my only point is that if you know the result for the unbounded case then the bounded case follows formally even though, as you point out, the model structure on one is not transferred from the model structure on the other. $\endgroup$ – Gabriel C. DrummondCole May 22 '15 at 14:47