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Consider $Y$ an element of the $n$-dimensional tangent space $T_yY$. The the canonical basis is $(\frac{\partial}{\partial y^1}, \cdots, \frac{\partial}{\partial y^n}).$ Then should I write $$Y = (Y^1\frac{\partial}{\partial y^1}, \cdots, Y^n\frac{\partial}{\partial y^n}),$$ or $$Y = \sum_{j=1}^n Y^j\frac{\partial}{\partial y^j}?$$

Or, as a matter of fact, these two notations are equivalent?

WishingFish
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3 Answers3

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Since $\frac{\partial}{\partial y^i}$ denotes an element of the basis of $T_yM$, your first notation doesn't make much sense. The second notation is the correct one. Also note that very often the Einstein summation convention is used, so you can omit the sum and write $$Y = Y^i\frac{\partial}{\partial y^i}$$

Daniel Robert-Nicoud
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  • Thanks Daniel. The problem is, I need to compute $f^\omega(Y)$, but the definition is given by $$f^w(v_1, \dots, v_p) = w(fv_1, \dots, fv_p).$$ So I want to write $Y$ as a $n$-tuple. – WishingFish Aug 05 '13 at 21:20
  • @WishingFish I think you have misunderstood what $f^*\omega(Y)$ means. If $\omega$ is a $1$-form, then it will take $1$ vector. – Daniel Robert-Nicoud Aug 05 '13 at 21:23
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It is usual to write $$Y^1 \tfrac{\partial}{\partial y_1} + \cdots + Y^n \tfrac{\partial}{\partial y_n}$$

Each of the $\partial/\partial y_i$, or $\partial_i$ for short, are tangent vector fields on your manifold. At a given point each of the $\partial_i$ are tangent vectors. Provided that your manifold is regular, each of the $\partial_i$ will be linearly independent tangent vectors, and so $\partial_1,\ldots,\partial_n$ form a basis for the tangent plane.

Each tangent vector in a tangent plane can be writen as a linear combination of the tangent vectors $\partial_1,\ldots,\partial_n$. Hence $Y = (Y^1,\ldots,Y^n)$ with respect to the basis $(\partial_1,\ldots,\partial_n)$.

In what follows, we much change our notation.

Let's say you have two manifolds $X$ and $Y$, and your have a differentiable function $\operatorname{f} : X \to Y$. Let $\omega$ be a differentiable $k$-form on $Y$. The pull-back of $\omega$, under $\operatorname{f}$, denoted by $\operatorname{f}^*\!\omega$ is a differentiable $k$-form on $X$ $$(\operatorname{f}^*\!\omega)(v_1,\dots,v_k) := \omega(\operatorname{f}_*\!v_1,\ldots,\operatorname{f}_*\!v_k)$$ where $\operatorname{f}_* : TX \to TY$. If you take local coordinates then the differential $\operatorname{f}_*$ has the Jacobian matrix as its matrix representation. Each of the $v_i$ is a tangent vector to $X$. Use the Jacobian matrix to transform these into $\operatorname{f}_*\!v_i$, which are tangent vectors to $Y$. Then evaluate the differential $k$-form each of these $\operatorname{f}_*\!v_i$ to get $(\operatorname{f}^*\!\omega)(v_1,\ldots,v_k).$

Fly by Night
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I would and do definitely favor

$Y = \sum_{j = 1}^n Y^j \frac{\partial}{\partial y^j}$

over

$Y = (Y^1\frac{\partial}{\partial y^1}, Y^2\frac{\partial}{\partial y^2}, . . ., Y^n\frac{\partial}{\partial y^n})$

for the vector $Y$. My reason is similar to Daniel Robert-Nicoud's: the only way I know how to make sense of the right-hand side of the equation $Y = (Y^1\frac{\partial}{\partial y^1}, Y^2\frac{\partial}{\partial y^2}, . . ., Y^n\frac{\partial}{\partial y^n})$ is as a matrix whose columns are $Y^k\frac{\partial}{\partial y^k}$, viz.:

$\begin{pmatrix} Y^1 & 0 & . . . & 0 \\ 0 & Y^2 & . . . & 0 \\ 0 & 0 & . . . & 0 \\ . \\ . \\ . \\ 0 & 0 & . . . & Y^n \end{pmatrix}$,

but the more usual column form for $Y$ is

$\begin{pmatrix} Y^1 \\ Y^2 \\ . \\ . \\ . \\ Y^n \end{pmatrix}$.

Since we reserve square matrices to represent linear maps, using a matrix for $Y$ is bound to cause notational confusion and inconsistency.

Robert Lewis
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