# I Uniqueness of tangent space at a point

1. Jul 12, 2018

### kent davidge

How do you show that there can be only one tangent space at a given point of a manifold? Geometrically it's pretty obvious in 3 dimensions, as one notices that there can be only one tangent plane at a point. But how could we show that using equations?

2. Jul 12, 2018

### Staff: Mentor

What is the definition of a tangent space at a given point in the manifold?

Start with that and suppose that there exists more than one what would it mean?

3. Jul 12, 2018

### mathwonk

actually the way you stated it it may be hard to show, since there are many different definitions of abstract tangent space. even if you give a definition of a tangent space, all your questions asks is whether that definition is well defined.

interestingly, uniqueness may be more closely related to the definition of manifold, since for say a figure eight curve, there are two tangent lines at the point where the branches of the eight cross. think about that.

4. Jul 12, 2018

### kent davidge

yes, but in this case the two tangent lines correspond to the two distinct points of the line which go to the same point in the immersion. So we still have the relation: one point - one tangent space. No?
The definition I'm aware of has to do with the differential map of the function which maps the points of the manifold to $\mathbb{R}^n$, but as @mathwonk pointed out there are more than one definition, so I think I will try to figure out which one is the easiest for me to use.

5. Jul 13, 2018

### lavinia

What definition is that? Can you define this precisely?

Abstractly tangent spaces are defined on smooth manifolds via operators on differentiable functions. The space of operators at each point is a vector space. It can be shown that the conditions that define these operators determines the tangent space uniquely.

Last edited: Jul 15, 2018
6. Jul 13, 2018

### kent davidge

Sure. The one I remember is:
- Consider a function $f: \mathcal U \rightarrow \mathbb{R}^n$, where $\mathcal U \subseteq \mathcal M$ of a topological manifold $\mathcal M$.
- Suppose that $y \in \mathcal U$ and that $f(0) = x$. The tangent space at $x$ is defined as all linear combinations of the form $$\alpha^\mu \frac{df}{dy^\mu}\bigg|_{y=0}$$ where $\alpha^\mu \in \mathbb R$ and $\alpha$ is a vector on the tangent space at $0$. So each such sum above with a different set of $\alpha^\mu$ comprises a vector on the tangent space at $x$.

I'm not sure if it matches with your definition

7. Jul 13, 2018

### mathwonk

if i were doing it i would try to show something like this: if f:U-->M is a diffeomorphism from an open neighborhood of 0 U in R^n to an n manifold M, taking 0 to x, then the differential of f takes R^n isomorphically to the tangent space of M at x. That to me is a uniqueness statement.

8. Jul 13, 2018

### kent davidge

In what sense?

9. Jul 13, 2018

### mathwonk

well the tangent space to R^n at any point is uniquely R^n, and this says that any chart carries this one unique model tangent space isomorphically to yours. so yours is determined by this. i.e. the unique tangent space to your manifold is whatever space R^n maps to by (the derivative of) any chart.

It is really not clear what is being asked for here. In calculus we define the tangent line as a certain limit, provided that limiting line is unique, i.e. the slope is defined as a certain limit provided that limit "exists", which means it is the unique accumulation point of our secant slopes. .so in calculus, tangent lines are unique by definition.

If you allow any limiting value of such secant slopes then for a figure eight curve you can get more than one tangent line, and the conclusion is that your figure eight curve is not a smooth manifold at the point. so in some sense, an embedded variety is a manifold if and only if its tangent space is unique.

if you consider an abstract manifold, then it has a definition for its tangent space, and if you accept that definition and if that definition specifies only one space, then again uniqueness is part of the definition. But as stated, the tangent space is not unique since the definitions are not unique. i.e. there are many equivalent definitions of a tangent space, and so uniqueness can only mean that they are compatible in the sense that any diffeomorphism of manifolds maps one definition to the other.

i.e. a tangent bundle construction is a functor that assigns to each manifold a vector bundle with certain properties: namely the tangent bundle of R^n is R^n x R^n, and any smooth map of manifolds induces a bundle map, and in the case of R^n that bundle map is given by the matrix of partial derivatives, and given any open subset of a manifold, the tangent bundle of that subset is the restriction to that subset of the tangent bundle on the big manifold. If you look in Spivak's differential geometry book vol 1, you will see this abstract functorial development characterizing the tangent bundle construction.

so before this question can be answered you have to make the question precise somehow. but for subsets of the plane, say plane curves, the usual definition makes sense and it is interesting to characterize those curves and those points at which the tangent line is in fact unique, and this is equivalent to plane curves which are either smooth manifolds at the point, or more generally have only one "branch" at the given point. e.g. the curve y^2 = x^3 is not a manifold at the origin but there is still only one tangent line there, the x axis. nonetheless, with the correct definition, (not just limiting positions of lines obtained from secants, but as the zero locus of the linear term of the defining polynomial), the tangent space at that point is actually 2 dimensional, which implies the curve is not a smooth submanifold of the plane at that point.

Last edited: Jul 15, 2018
10. Jul 13, 2018

### mathwonk

so you can view your manifold as embedded in R^n and then the question to me would be identifying which lines in R^n are tangent to your manifold, i.e. how your tangent bundle is identified as a subbundle of R^n x R^n. e.g. if your manifold is the zero locus of a smooth function, then the tangent space at a point is the zero locus of the differential of that function, provided that differential is not identically zero. It is this last condition that guarantees your zero locus is a manifold. or you can define tangent lines as limits of secants, or you can go abstract and define tangent vectors as derivations on the germ of smooth functions defined near your point. This is the one Lavinia mentioned and is a favorite, e.g. with John Lee, since it easily gives the vector space structure, which is not so obvious with the limits of secants definition. you can also use equivalence classes of curves through the point, essentially the same as tangential lines, or you can go more abstract and define a bundle by patching bundles on open sets of R^n via charts and their differentials. but whatever approach you take, you want a way to assign a ("unique") vector space to each point, in such a way that every smooth map of manifolds induces a family of linear maps of tangent spaces, that become identified with the usual differential on euclidean space via any chart. and compositions should behave well. there are thus many ways to define such a bundle, but any two ways will be mapped to each other isomorphically by the induced family of linear maps associated to the identity map of the manifold. that's what i mean when i say a uniqueness statement merely says that any one definition maps isomorphically to any other via a diffeomorphism. so as usual, given the axioms, uniqueness means there is a unique such thing up to unique isomorphism.

Last edited: Jul 13, 2018
11. Jul 14, 2018

### kent davidge

Thank you. Very informative posts. I will see if any questions still arise after thinking about what you said.

12. Jul 15, 2018

### mathwonk

to repeat, one might wonder first if you are asking whether for a manifold there is only one unique way to define a tangent space, and the answer to that is no, but each way to define it does give a unique space, and all ways are compatible via the derivative map. then one might also wonder whether you are asking for which spaces, not necessarily manifolds, the tangent space is still defined and unique. Then it turns out one can define a tangent space for any variety defined by polynomial equations, and then the tangent space is unique and of the correct dimension if and only if that algebraic variety is in fact a manifold. there do exist however "singular" algebraic varieties which have either tangent spaces of the wrong dimension (too large), or that have at a given singular point more than one tangent of the right dimension (such as the figure eight curve). I.e. the figure eight curve is not a manifold at the origin because it has two distinct tangent lines there, and the cusp curve y^2 = x^3 is not a manifold at the origin even though it has one tangent line there because with the algebraic definition of tangent space it has a 2 dimensional tangent space at the origin.

technically speaking an algebraic variety has two tangent objects at each point, a tangent cone and a tangent space, and these are equal if and onloy if the variety is a manifold at the given point. For the figure eight curve, th tangent cone at the origin is the union of the two distinct tangent lines, and the tangent space is the whole plane. for the cusp curve, the tangent cone is the x axis but counted twice, and the tangent space is the whole plane. there is nice free book on algebraic curves available on bill fulton's website at michigan, but requiring some commutative algebra.

http://www.math.lsa.umich.edu/~wfulton/CurveBook.pdf

see page 31 ff. he discusses curves that have singular points with multiple tangent lines, but does not discuss tangent "spaces" since since for all plane curves with a singularity the tangent space at the singularity is the whole plane hence gives no information. it is better to study the tangent lines. i.e. at a singular point the tangent "space" is too big, and it is better to study the tangent cone.

13. Jul 15, 2018

### lavinia

I think the definition you mean is all linear combinations $X=∑_{i}a_{i}∂/∂x_{i}$ of the coordinate vectors $∂/∂x_{i}$. Abstractly these combinations represent derivations at each point $p$. That is, they are real number valued operators on functions that satisfy two properties.

1) $X$ is linear. This means $X(af+bg)=aXf+bXg$ for any constants $a$ and $b$ and any two functions $f$ and $g$.
2)$Xfg = f(p)Xg+g(p)Xf$. This is called the Leibniz rule.

These two properties can be taken to be the definition of a tangent vector. Any derivation at $p$ is of the form $X=∑_{i}a_{i}∂/∂x_{i}$.

-One might wonder how derivations correspond to geometric tangent vectors to manifolds embedded in Euclidean space. These two definitions seem different. If one takes a curve $c(t)$ on the manifold whose derivative at $p$, $dc/dt$ at time $t$, equals a tangent vector $v$ then the operation $f→df(c(t))/dt$ at time $t$ is a derivation at $p$. This definition does not depend on the curve but only on $v$. On the other hand, every derivation at $p$ can be obtained as a derivative along a curve so every derivation corresponds to a geometric tangent vector.

- Every derivation at a point $p$ is a linear combination of partial derivative operators.
If ${x_{i}}$ are coordinates then a derivation $X$ is completely determined by the $n$ numbers $Xx_{i}$. So the vector space of derivations at $p$ is $n$ dimensional as is the subspace of linear combinations of partial derivative operators $∑_{i}a_{i}∂/∂x_{i}$. So they must be equal. This is why the tangent space can be defined as the space of derivations.

Last edited: Jul 16, 2018
14. Jul 16, 2018

### lavinia

Continuing with post #13:

In order to define a derivation, one only needs a differentiable manifold. There is no need for a coordinate system or a geometric realization. Derivations are intrinsic. In this sense they are uniquely determined.

It seems to me that geometric tangent vectors are dependent on the particular embedding or immersion and are not intrinsic to a manifold. A different embedding can give a different tangent plane at $F(p)$. However, the action of geometric tangent vectors as derivations is intrinsic. For instance suppose $F:M→R^{n}$ is an embedding and $v$ is tangent at $F(p)$. If $c(t)$ is a curve in $F(M)$ whose derivative equals $v$ at time $t$, then $f→ d(f \circ F^{-1} \circ c(t))/dt$ at time $t$ is a derivation at $p$.

Last edited: Jul 16, 2018
15. Jul 17, 2018

### kent davidge

Now you touched on a point which I was also thinking about. There are several ways of mapping a manifold into another. If we are to follow the geometric interpretation that the tangent plane at a point is obtained by following the mesh at the point, then of course it has to be different for different mappings. This conclusion is also clear from the definition given above, as the tangent vectors depend on the derivatives of the map, which are vectors in the target manifold (Here I'm thinking about $\mathbb{R}^n$ as the targent manifold, i.e., the codomain of the mapping.)
So should we differentiate between these two types?

16. Jul 18, 2018

### lavinia

All derivations come from differentiating along parameterized curves. One can in fact define tangent vectors as equivalence classes of parameterized curves where two curves are equivalent if differentiating along them at a point determines the same derivation.

The tangent vectors to a submanifold are those derivations on the larger ambient manifold which can be calculated along curves that lie completely on the submanifold. A curve is tangent if it is in the same equivalence class as a curve on the submanifold. Such a curve can wander off into the surrounding space. The tangent line to a surface in Euclidean space (when parameterized) is an example.

Last edited: Jul 19, 2018
17. Jul 19, 2018

### lavinia

Continuing with post #16

The trace of a curve i.e. its unparameterized path in the manifold, does not determine a derivation. Rather it determines a collection of derivations depending on how it is parameterized. If $c(t)$ and $c(t(s)$ are two parameterizations then $X_{c(s)}= dt/dsX_{c(t)}$. At the point $p$ the possible derivations trace a line through the origin of the tangent space.

Last edited: Jul 20, 2018