Let $V$ be a graded vector space over $\mathbb{k}$ and $V[1]$ its odd degree shift. Given $k$, $l\in \mathbb{N}_0$, is there a natural way to define the following map, $$ \psi: \hom_{\mathbb{k}}(V^{\otimes k},V^{\otimes l}) \longrightarrow \hom_{\mathbb{k}}(V[1]^{\otimes k},V[1]^{\otimes l}), $$ where $\hom_{\mathbb{k}}(V,W)$ denotes the graded vector space generated by $\mathbb{k}$-linear homogenous maps $V\rightarrow W$ of degree $d\in \mathbb{Z}$?
- The case $k=l=1$: Consider the dg-category of dg-vector spaces $\mathrm{dgVec}$ (objects are dg-vector spaces and morphisms degree $d$ chain maps). One defines the suspension $s$ on $\mathrm{dgVec}$ by mapping $V\mapsto V[1]$ and $(f: V \to W) \mapsto (\tilde{f}: V[1]\to W[1])$, where $\tilde{f}$ is defined with the help of the natural morphism $\theta_V\in \hom^1(V,V[1])$ as $$ \tilde{f}(\theta_V v) = (-1)^{|f|}\theta_W f(v) $$ for $v\in V$. This sign convention guarantees that $s$ is a dg-functor on $\mathrm{dgVec}$ and is hence a natural candidate for $\psi$ in this case.
- Idea from enriched categories: Equip $\mathrm{dgVec}$ with a (dg) symmetric mononoidal structure (note that there are two non-isomorphic by applying tensor products of morphisms either from the left to the right or the other way round and counting the Koszul sign). Is it possible to consider a dg-functor $s^{\otimes k}\otimes s^{\otimes l}$ of $\mathrm{dgVec}^{\otimes k}\otimes\mathrm{dgVec}^{\otimes l}$ and view $\psi$ as a kind of a dg-natural transformation? What conditions on $\psi$ would arise in this way? It is kind of weird because $\psi$ should have degree $\pm(l-k)$.
- Idea from operads: There is a notion of the degree shift of a dg-operad. It should be done such that $(\mathrm{end}_V)[1]$ is naturally isomorphic to $\mathrm{end}_{V[1]}$ as a dg-operad. I think that $\psi$ for $k\in\mathbb{N}_0$ and $l=1$ must be used here and must satisfy (up to some global signs) $$ \psi(f)(\theta v_1 \otimes \dotsb\theta v_k) = \varepsilon(\theta,v) \theta f(v_1\otimes \dotsb \otimes v_k),$$ where $\varepsilon(\theta,v)$ denotes the Koszul sign to reorder $\theta_1\dotsb\theta_k v_1\dotsb v_k$ to $\theta_1 v_1 \dotsb \theta_k v_k$. Extrapolating to properads, I would expect that the following definition is reasonable for a map $f:V^{\otimes k}\rightarrow V^{\otimes l}$: $$ \psi(f)(\theta^{\otimes k} v_1\otimes\dotsb\otimes v_k) = \theta^{\otimes l} f(v_1\otimes\dotsb\otimes v_k), $$ where the tensor product is evaluated with Koszul signs. I find the following claim for such $\psi$ intriguing: Given an algebra $V$ over a properad $P$, shifting the corresponding operations $f$ on $V$ using $\psi$ gives new operations $\psi(f)$ on $V[1]$ which make it into an algebra over $P[1]$.
- Idea from algebra with formal symbols: It seems to me that the following might be a reasonable definition: $$ \psi(f)(\theta^{\otimes k} v_1\otimes\dotsb\otimes v_k) = (-1)^{k |f| + \frac{1}{2} k(k-1)}\theta^{\otimes l}f(v_1\otimes\dotsb \otimes v_k). $$ It comes from writing $\psi(f) = \theta^{\otimes l}_* \bar{\theta}^{\otimes k *} f$, where $\bar{\theta}^{\otimes k *} f = (-1)^{k |f|} f\circ\bar{\theta}^{\otimes k}$ is a pullback (the sign must be there for compatibility with composition), where $\bar{\theta}$ is the desuspension, and the sign $(-1)^{\frac{1}{2}k(k-1)}$ comes from the "collision" $\bar{\theta}^{\otimes k}\theta^{\otimes k} = (-1)^{\frac{1}{2}k(k-1)} \mathrm{1}^{\otimes k}$. However, I did not see this definition anywhere and can not really argue why I should use all these signs.
What is your idea/reference/...? I know that it is not a strictly mathematical question, it is rather like if you were to choose $\psi$, how would you do it so that it would naturally fit into the most of the theories, where it appears. Thank you very much!
