503: Tangent Vectors Space at Point on C∞ Manifold with Boundary

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definition of tangent vectors space at point on $C^{\mathrm{\infty }}$ manifold with boundary

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Topics

About: $C^{\mathrm{\infty }}$ manifold

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The table of contents of this article

• Starting Context
• Target Context
• Orientation
• Main Body
• 1: Structured Description
• 2: Note

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Starting Context

• The reader knows a definition of tangent vector.
• The reader knows a definition of %field name% vectors space.

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Target Context

• The reader will have a definition of tangent vectors space at point on $C^{\mathrm{\infty }}$ manifold with boundary.

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Orientation

There is a list of definitions discussed so far in this site.

There is a list of propositions discussed so far in this site.

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Main Body

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1: Structured Description

Here is the rules of Structured Description.

Entities:
$M$: $\in \{\text{ the }{C}^{\mathrm{\infty }}\text{ manifolds with (possibly empty) boundary }\}$
$m$: $\in M$
$\ast T_{m}M$: $=\{\text{ the tangent vectors at }m\}$, $\in \{\text{ the }\mathbb{R}\text{ vectors spaces }\}$ with the operations specified below
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Conditions:
$\mathrm{\forall }r\in \mathbb{R},\mathrm{\forall }v\in T_{m}M,\mathrm{\forall }f\in C_m^{\mathrm{\infty }}(M)((rv)f=r(vf))$
$\wedge$
$\mathrm{\forall }{v}_1,v_2\in T_{m}M,\mathrm{\forall }f\in C_m^{\mathrm{\infty }}(M)((v_1+v_2)f=v_{1}f+v_{2}f)$
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2: Note

Let us see that $T_{m}M$ is indeed an $\mathbb{R}$ vectors space.

1st, let us see that the operations are well-defined.

Is $rv$ really a derivation?

For each $r^{\prime }\in \mathbb{R}$, $(rv)(r^{\prime }f)=r(v(r^{\prime }f))=r(r^{\prime }v(f))=(r{r}^{\prime })v(f)=(r^{\prime }r)v(f)=r^{\prime }(rv(f))=r^{\prime }((rv)(f))$, so, $rv$ is an $\mathbb{R}$-linear map.

$(rv)(f{f}^{\prime })=r(v(f{f}^{\prime }))=r(v(f)f^{\prime }+fv(f^{\prime }))=rv(f)f^{\prime }+rfv(f^{\prime })=(rv)(f)f^{\prime }+f(rv)(f^{\prime })$, so, $rv$ satisfies the Leibniz rule.

So, $rv$ is a derivation.

Is $v_1+v_2$ really a derivation?

For each $r^{\prime }\in \mathbb{R}$, $(v_1+v_2)(r^{\prime }f)=v_1(r^{\prime }f)+v_2(r^{\prime }f)=r^{\prime }{v}_1(f)+r^{\prime }{v}_2(f)=r^{\prime }(v_1(f)+v_2(f))=r^{\prime }(v_1+v_2)(f)$, so, $v_1+v_2$ is an $\mathbb{R}$-linear map.

$(v_1+v_2)(f{f}^{\prime })=v_1(f{f}^{\prime })+v_2(f{f}^{\prime })=v_1(f)f^{\prime }+f{v}_1(f^{\prime })+v_2(f)f^{\prime }+f{v}_2(f^{\prime })=(v_1(f)+v_2(f))f^{\prime }+f(v_1(f^{\prime })+v_2(f^{\prime }))=(v_1+v_2)(f)f^{\prime }+f(v_1+v_2)(f^{\prime })$, so, $v_1+v_2$ satisfies the Leibniz rule.

So, $v_1+v_2$ is a derivation.

1) for any elements, $v_1,v_2\in T_{m}M$, $v_1+v_2\in T_{m}M$ (closed-ness under addition): which has been seen above.

2) for any elements, $v_1,v_2\in T_{m}M$, $v_1+v_2=v_2+v_1$ (commutativity of addition): $(v_1+v_2)(f)=v_1(f)+v_2(f)=v_2(f)+v_1(f)=(v_2+v_1)(f)$.

3) for any elements, $v_1,v_2,v_3\in T_{m}M$, $(v_1+v_2)+v_3=v_1+(v_2+v_3)$ (associativity of additions): $((v_1+v_2)+v_3)(f)=(v_1+v_2)(f)+v_3(f)=v_1(f)+v_2(f)+v_3(f)=v_1(f)+(v_2(f)+v_3(f))=v_1(f)+(v_2+v_3)(f)=(v_1+(v_2+v_3))(f)$.

4) there is a 0 element, $0\in V$, such that for any $v\in T_{m}M$, $v+0=v$ (existence of 0 vector): the map that maps each $f$ to $0$, $0$, is obviously a derivation, so, $0\in T_{m}M$, and $(v+0)(f)=v(f)+0(f)=v(f)+0=v(f)$.

5) for any element, $v\in T_{m}M$, there is an inverse element, $v^{\prime }\in T_{m}M$, such that $v^{\prime }+v=0$ (existence of inverse vector): the map that maps each $f$ to $-v(f)$, $-v$, is obviously a derivation, so, $-v\in T_{m}M$, and $(v^{\prime }+v)(f)=(-v+v)(f)=-v(f)+v(f)=0=0(f)$.

6) for any element, $v\in T_{m}M$, and any scalar, $r\in \mathbb{R}$, $r.v\in T_{m}M$ (closed-ness under scalar multiplication): which has been seen above.

7) for any element, $v\in T_{m}M$, and any scalars, $r_1,r_2\in \mathbb{R}$, $(r_1+r_2).v=r_1.v+r_2.v$ (scalar multiplication distributability for scalars addition): $((r_1+r_2).v)(f)=(r_1+r_2)v(f)=r_{1}v(f)+r_{2}v(f)=(r_{1}v)(f)+(r_{2}v)(f)=(r_1.v+r_2.v)(f)$.

8) for any elements, $v_1,v_2\in T_{m}M$, and any scalar, $r\in \mathbb{R}$, $r.(v_1+v_2)=r.v_1+r.v_2$ (scalar multiplication distributability for vectors addition): $(r.(v_1+v_2))(f)=r(v_1+v_2)(f)=r(v_1(f)+v_2(f))=r{v}_1(f)+r{v}_2(f)=(r{v}_1)(f)+(r{v}_2)(f)=(r.v_1+r.v_2)(f)$.

9) for any element, $v\in T_{m}M$, and any scalars, $r_1,r_2\in \mathbb{R}$, $(r_1{r}_2).v=r_1.(r_2.v)$ (associativity of scalar multiplications): $((r_1{r}_2).v)(f)=(r_1{r}_2)v(f)=r_1(r_{2}v(f))=r_1((r_{2}v)(f))=(r_1.(r_2.v))(f)$.

10) for any element, $v\in T_{m}M$, $1.v=v$ (identity of 1 multiplication): $(1.v)(f)=1v(f)=v(f)$.

So, $T_{m}M$ is an $\mathbb{R}$ vectors space.

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References

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