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Is left invariant vector field on $\mathbb{G}$,is same as a vector field which is invariant under group transformation.

Nebo Alex
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1 Answers1

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I guess you need a plain english explanation.

A vector field $X$ is a function that associate smoothly to every point $p$ of $G$ an element or vector $X_p$ of the tangent space of the group $G$ (which in this case is also a manifold). So for every point $p$ you have the vector $X_p \in T_p(G)$.

Now suppose you have a function $F$ that maps every point of $G$ to another point of $G$. Let's suppose that the point $p$ goes to $q$ i.e. $F(p)=q$.

Then if you consider the vector field $X$ you already had, you can play two different games with this vector field:

  1. First of all you can calculate $X_{F(p)}$. This is the original vector field $X$ but evaluated in the new point $F(p)$ which is $q$, i.e. $X_{F(p)}=X_{q}$. In particular if you apply the vector to a function $f\in C^{\infty}(G)$, you have $X_{F(p)}(f)$;

  2. The other game you can do is to calculate $(F_*X_{p})(f)$ which by definition is $X_p(f\circ F)$. Here, $F_*X_{p}$ is given by taking the vector $X_p$ calculated from the old vector field and considering its image by a the linear application $F_*$ which is the differential of $F$.

The two games are in general different because in the first case you leave the vector field untouched and you change the point to which it is applied, while in the second case you use the old point you had and you evaluate the vector field in the point but then you move the resulting vector by applying the function.

Now if the two games coincide then you have that the vector field is invariant i.e. $$X_{F(p)}(f)=X_p(f\circ F)$$

You can think about this with a hypersimple physical example: Consider a river and a thermometer. The river is flowing so a particle of water is moved along the flow. If you're interested in the change of temperature along the flow of the river, at any given time you can either measure the temperature few istants after the time given measuring the temperature in the new place where the particles will be or you can measure now the temperature of the particles and assign this temperature to the place where you guess they will be given the velocity they have.

Now your case: if the function $F$ is the function given by the left multiplication of the group, then you have a "Left Invariant Vector Field".
Actually it's easier to work out some example to get effectively the idea.

Wurstcake
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Dac0
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    I had high hopes of understanding left invariant vector fields, right up until the point where we apply a vector to a function $f$. What is this function, and what does that mean? Later, we apply $F$ to $f$, suggesting that $f$ is an element of $G$? – A_P Aug 19 '19 at 21:28
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    well the fact that maybe is not trivial to know if you didn't have rigorous differential geometry, is that you can think to vector as of derivations where $e_1,e_2, etc..$ are just the partial derivatives $d/dx_1, d/dx_2, etc...$. So if you apply a vector to a function you just derive with a directional derivative... – Dac0 Jul 03 '20 at 17:53
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    Some comments expanding the answer: Let $F:M\to M'$ be a diffeomorphism between smooth manifolds. For a given smooth vector field $X:M\to TM$, the pushforward of $X$ along $F$, denoted $F_X$, is defined as $F_X=dF\circ X\circ F^{-1}$, which is a vector field on $M'$ (here $dF:TM\to TM'$ is the global differential). One can check that $(\operatorname{id}M)X=X$ and that if $M\xrightarrow{F}M'\xrightarrow{G}M''$ are diffeomorphisms, then $(G\circ F)_=G_\circ F_$. This is the functoriality of the pushfoward.

    Now suppose $F:M\to M$ is a diffeomorphism of $M$ onto itself. A vector

     – Elías Guisado Villalgordo Dec 04 '22 at 11:23
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    field $X$ on $M$ is said to be $F$-invariant if $X$ is $F$-related with itself, $F_*X=X$, i.e., $X$ is a fixed point of the pushforward along $F$. Equivalently, $X\circ F=dF\circ X$. By the functoriality of the pushforward, it follows that a vector field $X$ on $M$ is $F$-invariant if and only if it is $F^{-1}$-invariant.

    Now, let $G$ denote a Lie group, and for $g\in G$, denote $L_g:G\to G$ to the multiplication on the left by $g$. This is a diffeomorphism with inverse $L_{g^{-1}}$. A vector field $X$ on $G$ is said then to be left invariant if it is $L_g$-invariant for every $g\in G$.

     – Elías Guisado Villalgordo Dec 04 '22 at 11:31
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    $\def\aut{\operatorname{Aut}}$Another rephrasing of left invariance: for a smooth manifold $M$, denote $\Gamma(TM)$ to the real vector space of vector fields on $M$, and denote $\aut(M)$ to the group of diffeomorphisms of $M$ into itself. The group $\aut(M)$ acts on $\Gamma(TM)$ by the pushforward, so the set of $F$-invariant vector fields on $M$ is $\Gamma(TM)^F$, the fixed points of $\Gamma(TM)$ under $F$. If $G$ is a Lie group, there is an injective group homomorphism \begin{align} L:G&\to\aut(G)\ g&\mapsto L_g. \end{align} (This is just the left canonical action of – Elías Guisado Villalgordo Dec 04 '22 at 12:03
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    $G$ on itself.) This way, $G$ acts on $\Gamma(TG)$ via $L$. The family of left invariant vector fields on $G$ equals $\bigcap_{g\in G}\Gamma(TG)^g=\Gamma(TG)^G$, the set of elements of $\Gamma(TG)$ fixed under the left action of $G$. – Elías Guisado Villalgordo Dec 04 '22 at 12:04