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1. • CommentRowNumber1.
   • CommentAuthorAndrew Stacey
   • CommentTimeJun 3rd 2011
   • PermaLink
   Author: Andrew Stacey
   Format: MarkdownItexFirst stab at [[propagating flows]] (highly tempted to put in a redirect for *propogating* flows). I wrote it without reference to either my article or Veroniques' in the hope that by being forced to look at it afresh, I'd get the argument right. I'm not convinced that I managed it so I'll need to polish it considerably.

   First stab at propagating flows (highly tempted to put in a redirect for propogating flows). I wrote it without reference to either my article or Veroniques’ in the hope that by being forced to look at it afresh, I’d get the argument right. I’m not convinced that I managed it so I’ll need to polish it considerably.

2. • CommentRowNumber2.
   • CommentAuthorUrs
   • CommentTimeJun 3rd 2011
   • PermaLink
   Author: Urs
   Format: MarkdownItexAndrew, in the definition you wrote > $\phi(v) = \pi(v)$. That does not seem to make sense and does not seem to be the right definition. We want $\phi(v) : t v \mapsto v$ for $t \in [0,1]$, I think. No?

   Andrew,

   in the definition you wrote

   $\phi(v) = \pi(v)$.

   That does not seem to make sense and does not seem to be the right definition. We want

   $\phi(v) : t v \mapsto v$

   for $t \in [0,1]$, I think. No?

3. • CommentRowNumber3.
   • CommentAuthorUrs
   • CommentTimeJun 3rd 2011
   • PermaLink
   Author: Urs
   Format: MarkdownItexAh, wait, you take it with values in diffeomorphisms. Just a second...

   Ah, wait, you take it with values in diffeomorphisms. Just a second…

4. • CommentRowNumber4.
   • CommentAuthorUrs
   • CommentTimeJun 3rd 2011
   • PermaLink
   Author: Urs
   Format: MarkdownItexSo I think you want $\phi(v) : v \mapsto \pi(v)$?

   So I think you want $\phi(v) : v \mapsto \pi(v)$?

5. • CommentRowNumber5.
   • CommentAuthorAndrew Stacey
   • CommentTimeJun 3rd 2011
   • PermaLink
   Author: Andrew Stacey
   Format: MarkdownItexYes to your last, except that now that I see it abstracted away from the page then I see that it still doesn't quite make sense. It should be $\phi(v)$ takes $v$ to the _zero vector_ in its fibre. So $\phi(v) \colon v \mapsto 0_{\pi(v)}$ would be more correct.

   Yes to your last, except that now that I see it abstracted away from the page then I see that it still doesn’t quite make sense. It should be $\phi(v)$ takes $v$ to the zero vector in its fibre. So $\phi(v) \colon v \mapsto 0_{\pi(v)}$ would be more correct.

6. • CommentRowNumber6.
   • CommentAuthorUrs
   • CommentTimeJun 3rd 2011
   • PermaLink
   Author: Urs
   Format: MarkdownItexWait, $\phi(v)$ is a map $E \to E$. So I think $v \mapsto \pi(v)$ is good.

   Wait, $\phi(v)$ is a map $E \to E$. So I think $v \mapsto \pi(v)$ is good.

7. • CommentRowNumber7.
   • CommentAuthorUrs
   • CommentTimeJun 3rd 2011
   • PermaLink
   Author: Urs
   Format: MarkdownItexDo we really want to require that $\phi(v)$ is a _diffeomorphism_ instead of just a smooth map?

   Do we really want to require that $\phi(v)$ is a diffeomorphism instead of just a smooth map?

8. • CommentRowNumber8.
   • CommentAuthorAndrew Stacey
   • CommentTimeJun 3rd 2011
   • PermaLink
   Author: Andrew Stacey
   Format: MarkdownItexFor #6: No, because $E$ is a vector bundle over $M$ and $\pi \colon E \to M$ is the projection map. So $\pi(v)$ is an element of $M$ and we need to compose with the zero section to get it back in to $E$. For #7: I guess that technically we don't need it to be a diffeomorphism, but it has to be the identity outside a compact set and when $v = 0$ then it has to be the identity. Wait ... no, it does need to be a diffeomorphism. We are going to use this to define, for example, a diffeomorphism $\Omega M \times \mathbb{R}^n \cong L_U M$ where $L_U M \coloneqq \{ \alpha \in L M : \alpha(1) \in U\}$ and $\mathbb{R}^n \xrightarrow{\cong} U$ is a chart on $M$ centred at the basepoint (I'll identify $U$ and $\mathbb{R}^n$ via this map to avoid adding too many maps to the notation). The map is defined by $(\alpha,v) \mapsto \phi(v)^{-1} \circ \alpha$ with inverse $\alpha \mapsto (\phi(\alpha(1)) \circ \alpha,\alpha(1))$. So we do need $\phi(v)$ to be a diffeomorphism.

   For #6: No, because $E$ is a vector bundle over $M$ and $\pi \colon E \to M$ is the projection map. So $\pi(v)$ is an element of $M$ and we need to compose with the zero section to get it back in to $E$.

   For #7: I guess that technically we don’t need it to be a diffeomorphism, but it has to be the identity outside a compact set and when $v = 0$ then it has to be the identity. Wait … no, it does need to be a diffeomorphism. We are going to use this to define, for example, a diffeomorphism $\Omega M \times \mathbb{R}^n \cong L_U M$ where $L_U M \coloneqq \{ \alpha \in L M : \alpha(1) \in U\}$ and $\mathbb{R}^n \xrightarrow{\cong} U$ is a chart on $M$ centred at the basepoint (I’ll identify $U$ and $\mathbb{R}^n$ via this map to avoid adding too many maps to the notation). The map is defined by $(\alpha,v) \mapsto \phi(v)^{-1} \circ \alpha$ with inverse $\alpha \mapsto (\phi(\alpha(1)) \circ \alpha,\alpha(1))$. So we do need $\phi(v)$ to be a diffeomorphism.

9. • CommentRowNumber9.
   • CommentAuthorUrs
   • CommentTimeJun 3rd 2011
   • PermaLink
   Author: Urs
   Format: MarkdownItexre 6, okay I see, I misunderstood you as saying you want the result to be a vector. But you really mean the point (of course in a vector bundle). re 7, okay, I see, thanks.

   re 6, okay I see, I misunderstood you as saying you want the result to be a vector. But you really mean the point (of course in a vector bundle).

   re 7, okay, I see, thanks.

10. • CommentRowNumber10.
    • CommentAuthorAndrew Stacey
    • CommentTimeJun 6th 2011
    • PermaLink
    Author: Andrew Stacey
    Format: MarkdownItexI thought of a way to construct the diffeomorphisms more directly, thus avoiding the need for the exponentiation from vector fields. It still uses the same basic idea, but since it is a bit more explicit it will work in places where the exponentiation can't be assumed to exist. (As is always the way, when one returns to something several years later then one sees a simpler method.) Comments osv[1] welcomed. [1] To forestall the irrelevancies, "og så videre" which translates to "et cetera".

    I thought of a way to construct the diffeomorphisms more directly, thus avoiding the need for the exponentiation from vector fields. It still uses the same basic idea, but since it is a bit more explicit it will work in places where the exponentiation can’t be assumed to exist.

    (As is always the way, when one returns to something several years later then one sees a simpler method.)

    Comments osv[1] welcomed.

    [1] To forestall the irrelevancies, “og så videre” which translates to “et cetera”.

11. • CommentRowNumber11.
    • CommentAuthorUrs
    • CommentTimeJun 6th 2011
    • PermaLink
    Author: Urs
    Format: MarkdownItex> I thought of And you have added it to the entry, I suppose? Thanks, I'll have a look.

    I thought of

    And you have added it to the entry, I suppose? Thanks, I’ll have a look.

12. • CommentRowNumber12.
    • CommentAuthorAndrew Stacey
    • CommentTimeJun 6th 2011
    • PermaLink
    Author: Andrew Stacey
    Format: MarkdownItexYes. Although "added" might be the wrong word. I took out the original construction (it would have needed considerable cleaning up anyway) and replaced it with this one. The constructions are fairly similar, it's just that the last step is made explicit rather than calling on the exponentiation map.

    Yes. Although “added” might be the wrong word. I took out the original construction (it would have needed considerable cleaning up anyway) and replaced it with this one. The constructions are fairly similar, it’s just that the last step is made explicit rather than calling on the exponentiation map.