ScalingStacks

Example 2.11. For $B=\BO(n)$, with the usual map $\BO(n)\rightarrow\BTop(n)$, the $\oo$-category of topological $n$-disks with $\BO(n)$-framings is equivalent to the $\oo$-category of smooth $n$-disks and smooth embeddings, $\disk_{n}^{\sm}\simeq\disk_{n}^{{\sf BO}(n)}$. These are both equivalent to the PROP associated to the unoriented version of the ribbon, or “framed,” $\mathcal{E}_{n}$ operad; see [SW] for a treatment of this operad.⁵⁵ 5 The historical use of “framed” here is potentially misleading, since in the “framed” $\mathcal{E}_{n}$ operad the embeddings do not preserve the framing, while in the usual $\mathcal{E}_{n}$ operad the embeddings do preserve the framing (up to scale). It might lead to less confusion to replace the term “framed $\mathcal{E}_{n}$ operad” with “unoriented $\mathcal{E}_{n}$ operad.” To see these equivalences it is enough to explain why each of the natural maps $\mathsf{O}(n)\rightarrow\Emb^{\sm}(\mathbb{R}^{n},\mathbb{R}^{n})\rightarrow\Emb^{{\sf BO}(n)}(\mathbb{R}^{n},\mathbb{R}^{n})$ is an equivalence. Smoothing theory ([KS]) gives the equivalence of the second map. Via Gram–Schmidt, the inclusion $\mathsf{O}(n)\xrightarrow{\simeq}{\sf GL}(\mathbb{R}^{n})$ is a deformation retraction. Conjugation by scaling and translation, $(f,t)\mapsto\bigl(x\mapsto\frac{f(tx)-f(0)}{t}+f(0)\bigr)$, demonstrates the inclusion ${\sf GL}(\mathbb{R}^{n})\xrightarrow{\simeq}\Emb^{\sf sm}(\mathbb{R}^{n},\mathbb{R}^{n})$ as a deformation retraction.