Sameer Satish Shah and Bipasha Roy are presently associated as directors. Sameer Satish.Ģ Directors are associated with the organization. It's a company limited by shares having an authorized capital of Rs 1.00 lakh and a paid-up capital of Rs 1.00 lakh as per MCA.Ģ Directors are associated with the organization. The Company's status is Active, and it has filed its Annual Returns and Financial Statements up to (FY 2020-2021). Its registered office is in Mumbai, Maharashtra, india. In order to understand what is going on, I suggest you draw a picture.Mobius Space Consulting Private Limited is a 6 years 6 months old Private Company incorporated on. This will result in overtwisted Moebius bands in $R^3$. Exhibiting Analogous Triply and Singly Twisted Mbius Topologies. You can also replace $g$ with its odd powers. The latter effect creates a particularly convenient access to Mbius aromatic molecules. If you glue $P_+, P_-$ using $g$ you obtain a Moebius band embedded in $R^3$ in a standard way. One can also use a "half-twist" $g$ instead of $f$ ( $g$ has the property that $g\circ g=f$: It is a 180 degree rotation inside the circle $\alpha$ in $P$ and swaps $p_1, q_1$ and fixes the midpoint of $p_1q_1$). I note only that if instead of gluing using $f$ you perform gluing using its $n$th power, you obtain an $n$-times overtwisted annulus in $R^3$. Proving that this is true will take some effort and I will not do it. However, under the homeomorphic identification $X\to R^3$ the annulus $A$ maps to a "once overtwisted annulus" (doubly twisted Moebius band that you are interested in). One can show that $X$ is homeomorphic to $R^3$ (for instance, because $f: P\to P$ is isotopic to the identity map). The annulus $A$ still sits inside $X$ (since $f$ was the identity at $A\cap P$). I will define a new topological space $X$ obtained by gluing $P_-$ to $P_+$ using the homeomorphism $f$ of their boundaries. Let $P_\pm$ denote the lower and upper half-spaces in $R^3$ bounded by the plane $P$. In particular, it is the identity on the intersection $A\cap P$. (For the record, there is a left and right Dehn twists, it does not matter for our purpose which one do we use.) This homeomorphism will be identity outside of a small neighborhood of the loop $\alpha$. The precise location of this circle is irrelevant.Ĭonsider the surface $P$ equal to the $xy$-plane and let a homeomorphism $f: P\to P$ be the Dehn twist along the loop $\alpha$. Now, draw a round circle $\alpha$ in the $xy$-plane which separates $p_1,q_1$ from $p_2,q_2$. The boundary of $A$ is the union of two round circles $C_1, C_2$, which I will assume to be concentric with the common center at the origin $(0,0,0)$ and respective radii 1 and 2. Start with $R^3$ and a "vertical round annulus" $A$ contained the $xz$-plane. (For the record: This is not the best way to get "doubly twisted Moebius bands": A better way is to used a " framed unknot": Say, a round circle $C$ equipped with a normal vector field that twists once around $C$.)
I will write down an answer but it might take you awhile to understand it. What you are asking is for a rigorous description of a "cut-and-paste operation on the Euclidean 3-space $R^3$ which produces the doubly twisted Moebius band in $R^3$." You know how to obtain surfaces (such as Moebius bands) by identifying sides of polygons. I think I (finally) understood your question. This also implies that they have different knot groups, which I know nothing of. Suppose we are given the space $^$ because when we add another twist to $(0,p)\sim(1,1-p)$, we again get $(0,p)\sim(1,p)$, but when we look at the edges of the two structures, the cylinder has two disconnected circles while the double Möbius has two linked circles, hence there is no continuous deformation of space so that the cylinder can be converted to this. A few days back, I was taught about equivalence relations, classes and was given some glimpses of quotient topology.
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