### 数学代写|交换代数代写commutative algebra代考|MATH3033

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## 数学代写|交换代数代写commutative algebra代考|Tangent Vectors and Derivations

A decisive example of a vector bundle is the tangent bundle, for which the elements are the pairs $(p, v)$ where $p \in V$ and $v$ is a tangent vector at the point $p$.

When the manifold $V$ is a manifold immersed in a space $\mathbb{R}^{n}$, a tangent vector $v$ at the point $p$ can be identified with the derivation at the point $p$ in the direction of $v$.
When the manifold $V$ is not a manifold immersed in a space $\mathbb{R}^{n}$, a tangent vector $v$ can be defined as a derivation at the point $p$, i.e. as an $\mathbb{R}$-linear form $v: \mathbf{A} \rightarrow \mathbb{R}$ which satisfies Leibniz’s rule
$$v(f g)=f(p) v(g)+g(p) v(f) .$$
We can check with a few computations that the tangent vectors at $V$ indeed form a vector bundle $\mathrm{T}_{V}$ over $V$.

To a vector bundle $\pi: W \rightarrow V$ is associated the $\mathbf{A}$-module $\Gamma(W)$ formed by the smooth sections of the bundle. In the tangent bundle case, $\Gamma\left(\mathrm{T}_{V}\right)$ is nothing else but the $\mathbf{A}$-module of the usual (smooth) vector fields.

Just as a tangent vector at the point $p$ is identified with a derivation at the point $p$, which can be defined in algebraic terms (Eq. (1)), a (smooth) tangent vector field can be identified with an element of the A-module of the derivations of the $\mathbb{R}$-algebra $\mathbf{A}$, defined as follows.

A derivation of an $\mathbb{R}$-algebra $\mathbf{B}$ in a $\mathbf{B}$-module $M$ is an $\mathbb{R}$-linear mapping $v: \mathbf{B} \rightarrow M$ which satisfies Leibniz’s rule
$$v(f g)=f v(g)+g v(f) .$$
The $\mathbf{B}$-module of derivations of $\mathbf{B}$ in $M$ is denoted by $\operatorname{Der}{\mathbb{R}}(\mathbf{B}, M)$. When we “simply” refer to a derivation of an $\mathbb{R}$-algebra $g B$, what we mean is a derivation with values in $\mathbf{B}$. When the context is clear we write $\operatorname{Der}(\mathbf{B})$ as an abbreviation for $\operatorname{Der}{\mathbb{R}}(\mathbf{B}, \mathbf{B})$.

## 数学代写|交换代数代写commutative algebra代考|Differentials and Cotangent Bundle

The dual bundle of the tangent bundle, called the cotangent bundle, has the differential forms on the manifold $V$ as its sections.

The corresponding $\mathbf{A}$-module, called the module of differentials, can be defined by generators and relations in the following way.

Generally, if $\left(f_{i}\right){i \in I}$ is a family of elements that generate an $\mathbb{R}$-algebra $\mathbf{B}$, the $\mathbf{B}-$ module of (Kähler) differentials of $\mathbf{B}$, denoted by $\Omega{\mathbf{B} / \mathbb{R}}$, is generated by the (purely formal) $\mathrm{d} f_{i}$ ‘s subject to the relations “derived from” the relations that bind the $f_{i}$ ‘s: if $P \in \mathbb{R}\left[z_{1}, \ldots, z_{n}\right]$ and if $P\left(f_{i_{1}}, \ldots, f_{i_{n}}\right)=0$, the derived relation is
$$\sum_{k=1}^{n} \frac{\partial P}{\partial z_{k}}\left(f_{i_{1}}, \ldots, f_{i_{n}}\right) \mathrm{d} f_{i_{k}}=0$$
Furthermore, we have the canonical mapping $\mathrm{d}: \mathbf{B} \rightarrow \Omega_{\mathbf{B} / \mathbb{R}}$ available, defined by $\mathrm{d} f=$ the class of $f$ (if $f=\sum \alpha_{i} f_{i}$, with $\alpha_{i} \in \mathbb{R}, \mathrm{d} f=\sum \alpha_{i} \mathrm{~d} f_{i}$ ), which is a derivation. ${ }^{1}$

We then prove that, for every $\mathbb{R}$-algebra $\mathbf{B}$, the $\mathbf{B}$-module of derivations of $\mathbf{B}$ is the dual module of the $\mathbf{B}$-module of Kähler differentials.

In the case where the $\mathbf{B}$-module of differentials of $\mathbf{B}$ is a finitely generated projective module (for example when $\mathbf{B}=\mathbf{A}$ ), then it is itself the dual module of the $\mathbf{B}$-module of derivations of $\mathbf{B}$.

## 数学代写|交换代数代写commutative algebra代考|Tangent Vectors and Derivations

$$v(f g)=f(p) v(g)+g(p) v(f) .$$

$$v(f g)=f v(g)+g v(f) .$$

## 数学代写|交换代数代写commutative algebra代考|Differentials and Cotangent Bundle

$$\sum_{k=1}^{n} \frac{\partial P}{\partial z_{k}}\left(f_{i_{1}}, \ldots, f_{i_{n}}\right) \mathrm{d} f_{i_{k}}=0$$

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