假设我们现在有域名<你的域名>，我们可以使用DNS提供商（以Cloudflare为例）的API定期更新IPv6。




1. DNS添加AAAA记录(指向你的IPv6)

登录[https://dash.cloudflare.com](https://dash.cloudflare.com)，在Cloudflare中加入域，并将你的域名的DNS解析服务器（你购买域名的服务商）替换成Cloudflare的DNS服务器，在该域名的DNS管理后台添加记录：
```
类型      名称     IPv6地址       代理状态     TTL
AAAA     www      <随便填一个>    仅DNS       2分钟
```
并在dash页面的该域名下的概述界面的右下角，获取您的API令牌，并在此处记住你的域名的区域ID（ZONE ID）。

2. 安装DDNS脚本
安装必要的软件（Debian系下）：
```
sudo apt install curl jq
```
创建一个如下脚本`/usr/local/bin/ddns-ipv6.sh`，注意修改配置部分的内容:
```Shell
#!/bin/bash

LOGFILE="/var/log/ddns-ipv6.log"

touch $LOGFILE
chmod 644 $LOGFILE

echo "=== $(date) Starting DDNS update ===" >> $LOGFILE

# 配置部分
CF_API_TOKEN="你的Cloudflare API Token"
CF_ZONE_ID="你的Zone ID"
RECORD_NAME="你的域名"
NET_INTERFACE="enp0s31f6"  # 改为你的实际网卡名
TTL=120

# 日志函数
log() {
    echo "[$(date '+%Y-%m-%d %H:%M:%S')] $1"
}

# 尝试多种方法获取可靠的公网IPv6
get_ipv6() {
    # 方法1: 使用系统命令（推荐获取主要地址）
    local ip_system=$(ip -6 addr show dev $NET_INTERFACE scope global | grep -E 'global (dynamic|mngtmpaddr)' | awk '/inet6/{print $2}' | cut -d'/' -f1 | head -1)
    
    # 方法2: 使用公共API（备用）
    local ip_api=$(curl -6 --connect-timeout 5 -s https://api64.ipify.org || curl -6 --connect-timeout 5 -s https://ipv6.icanhazip.com)
    
    # 优先使用系统检测的IP，其次使用API检测的
    [ -n "$ip_system" ] && echo $ip_system || echo $ip_api
}

# 获取当前IPV6
IP=$(get_ipv6)

if [ -z "$IP" ]; then
    log "错误：无法获取有效的IPv6地址"
    exit 1
fi

log "检测到的IPv6地址: $IP"

# 获取记录ID
RESPONSE=$(curl -s -X GET "https://api.cloudflare.com/client/v4/zones/$CF_ZONE_ID/dns_records?type=AAAA&name=$RECORD_NAME" \
    -H "Authorization: Bearer $CF_API_TOKEN" \
    -H "Content-Type: application/json")

# 检查是否成功
SUCCESS=$(echo "$RESPONSE" | jq -r '.success')
if [ "$SUCCESS" != "true" ]; then
    ERRORS=$(echo "$RESPONSE" | jq -r '.errors[]?.message')
    log "错误：获取记录ID失败 - $ERRORS"
    exit 1
fi

RECORD_ID=$(echo "$RESPONSE" | jq -r '.result[0].id')

if [ -z "$RECORD_ID" ] || [ "$RECORD_ID" = "null" ]; then
    log "错误：未找到记录 $RECORD_NAME，请先在Cloudflare创建此AAAA记录"
    exit 1
fi

log "找到记录ID: $RECORD_ID"

# 更新DNS记录
UPDATE_RESPONSE=$(curl -s -X PUT "https://api.cloudflare.com/client/v4/zones/$CF_ZONE_ID/dns_records/$RECORD_ID" \
    -H "Authorization: Bearer $CF_API_TOKEN" \
    -H "Content-Type: application/json" \
    --data "{\"type\":\"AAAA\",\"name\":\"$RECORD_NAME\",\"content\":\"$IP\",\"ttl\":$TTL,\"proxied\":false}")

# 检查更新结果
UPDATE_SUCCESS=$(echo "$UPDATE_RESPONSE" | jq -r '.success')
if [ "$UPDATE_SUCCESS" = "true" ]; then
    NEW_CONTENT=$(echo "$UPDATE_RESPONSE" | jq -r '.result.content')
    log "成功更新记录! $RECORD_NAME → $NEW_CONTENT"
else
    ERRORS=$(echo "$UPDATE_RESPONSE" | jq -r '.errors[]?.message')
    log "错误：更新记录失败 - $ERRORS"
    exit 1
fi

echo "=== $(date) Finished DDNS update ===" >> $LOGFILE
```
并添加权限
```
chmod +x /usr/local/bin/ddns-ipv6.sh
```

3. 使用crontab定时任务自动更新
```
crontab -e
```
加入
```
*/5 * * * * /usr/local/bin/ddns-ipv6/sh > /dev/null 2>&1
```
这样每5分钟检测一次IPv6变化并更新DNS。

DDNS配置IPv6

# 张量积 II

## 1. 简介

继续研究张量乘积，我们将看到如何把两个线性映射 $M \rightarrow M^{\prime}$ 和 $N \rightarrow N^{\prime}$ 结合成一个线性映射$M \otimes_{R} N \rightarrow M^{\prime} \otimes_{R} N^{\prime}$。这就引出了平坦模块和基扩张之间的线性映射。然后，我们将研究向量空间的张量积（包括缩并）的特殊性，$R$-代数的张量积以及最后的$R$-模的张量代数。



## 2. 线性映射中的张量积

## 3. 平坦模

## 4. 线性映射和基扩张的张量积

## 5. 向量空间

## 6. 张量缩并

## 7. $R$-代数的张量积

## 8. $R$-模的张量代数

张量积 II

张量积 I

# Appendix
## A.1 Matrix Inversion
**Exercise A.1.1** Using the Method described above, show that
$$
\begin{pmatrix}
2 & 1 & 3\\
0 & 1 & 2\\
-1 & 1 & 1
\end{pmatrix}^{-1}=\begin{pmatrix}
1 & -2 & 1\\
2 & -5 & 4\\
-1 & 3 & -2
\end{pmatrix}
$$
and
$$
\begin{pmatrix}
2 & 1 & 3\\
4 & 1 & 2\\
0 & -1 & 2
\end{pmatrix}^{-1}=\dfrac{1}{12}
\begin{pmatrix}
-4 & 5 & 1\\
8 & -4 & -8\\
4 & -2 & 2
\end{pmatrix}
$$

**Solution.** (1) For given matrix
$$
M = \begin{pmatrix}
2 & 1 & 3 \\
0 & 1 & 2 \\
-1 & 1 & 1
\end{pmatrix}
$$
Define the row vectors:
$$
\mathbf{A} = (2, 1, 3),\quad \mathbf{B} = (0, 1, 2),\quad \mathbf{C} = (-1, 1, 1)
$$
Compute the reciprocal vectors
$$
\begin{align*}
\mathbf{A}_R &= \mathbf{B} \times \mathbf{C} = 
\begin{vmatrix}
\mathbf{i} & \mathbf{j} & \mathbf{k} \\
0 & 1 & 2 \\
-1 & 1 & 1
\end{vmatrix} \\
&= \mathbf{i}(1 \cdot 1 - 2 \cdot 1) - \mathbf{j}(0 \cdot 1 - 2 \cdot (-1)) + \mathbf{k}(0 \cdot 1 - 1 \cdot (-1)) \\
&= \mathbf{i}(-1) - \mathbf{j}(2) + \mathbf{k}(1) = (-1, -2, 1)
\end{align*}
$$
$$
\begin{align*}
\mathbf{B}_R &= \mathbf{C} \times \mathbf{A} = 
\begin{vmatrix}
\mathbf{i} & \mathbf{j} & \mathbf{k} \\
-1 & 1 & 1 \\
2 & 1 & 3
\end{vmatrix} \\
&= \mathbf{i}(1 \cdot 3 - 1 \cdot 1) - \mathbf{j}(-1 \cdot 3 - 1 \cdot 2) + \mathbf{k}(-1 \cdot 1 - 1 \cdot 2) \\
&= \mathbf{i}(2) - \mathbf{j}(-5) + \mathbf{k}(-3) = (2, 5, -3)
\end{align*}
$$
$$
\begin{align*}
\mathbf{C}_R &= \mathbf{A} \times \mathbf{B} = 
\begin{vmatrix}
\mathbf{i} & \mathbf{j} & \mathbf{k} \\
2 & 1 & 3 \\
0 & 1 & 2
\end{vmatrix} \\
&= \mathbf{i}(1 \cdot 2 - 3 \cdot 1) - \mathbf{j}(2 \cdot 2 - 3 \cdot 0) + \mathbf{k}(2 \cdot 1 - 1 \cdot 0) \\
&= \mathbf{i}(-1) - \mathbf{j}(4) + \mathbf{k}(2) = (-1, -4, 2)
\end{align*}
$$
Construct the cofactor matrix
$$
\overline{M} = \begin{pmatrix}
-1 & 2 & -1 \\
-2 & 5 & -4 \\
1 & -3 & 2
\end{pmatrix}
$$
Compute the determinant
$$
\begin{align*}
\det M &= \mathbf{A} \cdot (\mathbf{B} \times \mathbf{C}) \\
&= \begin{vmatrix}
2 & 1 & 3 \\
0 & 1 & 2 \\
-1 & 1 & 1
\end{vmatrix} \\
&= 2(1 \cdot 1 - 2 \cdot 1) - 1(0 \cdot 1 - 2 \cdot (-1)) + 3(0 \cdot 1 - 1 \cdot (-1)) \\
&= 2(-1) - 1(2) + 3(1) = -2 - 2 + 3 = -1
\end{align*}
$$
The inverse matrix is
$$
M^{-1} = \frac{\overline{M}}{\det M} = \frac{1}{-1} \begin{pmatrix}
-1 & 2 & -1 \\
-2 & 5 & -4 \\
1 & -3 & 2
\end{pmatrix} = \begin{pmatrix}
1 & -2 & 1 \\
2 & -5 & 4 \\
-1 & 3 & -2
\end{pmatrix}
$$
(2) For given matrix
$$
M = \begin{pmatrix}
2 & 1 & 3 \\
4 & 1 & 2 \\
0 & -1 & 2
\end{pmatrix}
$$
Define the row vectors as
$$
\mathbf{A} = (2, 1, 3),\quad \mathbf{B} = (4, 1, 2),\quad \mathbf{C} = (0, -1, 2)
$$
Compute the reciprocal vectors
$$
\begin{align*}
\mathbf{A}_R &= \mathbf{B} \times \mathbf{C} = 
\begin{vmatrix}
\mathbf{i} & \mathbf{j} & \mathbf{k} \\
4 & 1 & 2 \\
0 & -1 & 2
\end{vmatrix} \\
&= \mathbf{i}(1 \cdot 2 - 2 \cdot (-1)) - \mathbf{j}(4 \cdot 2 - 2 \cdot 0) + \mathbf{k}(4 \cdot (-1) - 1 \cdot 0) \\
&= \mathbf{i}(4) - \mathbf{j}(8) + \mathbf{k}(-4) = (4, -8, -4)
\end{align*}
$$
$$
\begin{align*}
\mathbf{B}_R &= \mathbf{C} \times \mathbf{A} = 
\begin{vmatrix}
\mathbf{i} & \mathbf{j} & \mathbf{k} \\
0 & -1 & 2 \\
2 & 1 & 3
\end{vmatrix} \\
&= \mathbf{i}((-1) \cdot 3 - 2 \cdot 1) - \mathbf{j}(0 \cdot 3 - 2 \cdot 2) + \mathbf{k}(0 \cdot 1 - (-1) \cdot 2) \\
&= \mathbf{i}(-5) - \mathbf{j}(-4) + \mathbf{k}(2) = (-5, 4, 2)
\end{align*}
$$
$$
\begin{align*}
\mathbf{C}_R &= \mathbf{A} \times \mathbf{B} = 
\begin{vmatrix}
\mathbf{i} & \mathbf{j} & \mathbf{k} \\
2 & 1 & 3 \\
4 & 1 & 2
\end{vmatrix} \\
&= \mathbf{i}(1 \cdot 2 - 3 \cdot 1) - \mathbf{j}(2 \cdot 2 - 3 \cdot 4) + \mathbf{k}(2 \cdot 1 - 1 \cdot 4) \\
&= \mathbf{i}(-1) - \mathbf{j}(-8) + \mathbf{k}(-2) = (-1, 8, -2)
\end{align*}
$$
Construct the cofactor matrix:
$$
\overline{M} = \begin{pmatrix}
4 & -5 & -1 \\
-8 & 4 & 8 \\
-4 & 2 & -2
\end{pmatrix}
$$
The determinant is
$$
\begin{align*}
\det M &= \mathbf{A} \cdot (\mathbf{B} \times \mathbf{C}) \\
&= \begin{vmatrix}
2 & 1 & 3 \\
4 & 1 & 2 \\
0 & -1 & 2
\end{vmatrix} \\
&= 2(1 \cdot 2 - 2 \cdot (-1)) - 1(4 \cdot 2 - 2 \cdot 0) + 3(4 \cdot (-1) - 1 \cdot 0) \\
&= 2(4) - 1(8) + 3(-4) = 8 - 8 - 12 = -12
\end{align*}
$$
Inverse matrix is 
$$
M^{-1} = \frac{\overline{M}}{\det M} = \frac{1}{-12} \begin{pmatrix}
4 & -5 & -1 \\
-8 & 4 & 8 \\
-4 & 2 & -2
\end{pmatrix} = \frac{1}{12} \begin{pmatrix}
-4 & 5 & 1 \\
8 & -4 & -8 \\
4 & -2 & 2
\end{pmatrix}
$$
$$
~\tag*{$\blacksquare$}
$$
## A.2 Gaussian Integrals
## A.3 Complex Numbers
## A.4 The $\mathrm{i}\varepsilon$ Prescription

Appendix

# Chapter 21 Path Integrals——II
## 21.1 Derivation of the Path Integral
## 21.2 Imaginary Time Formalism
## 21.3 Spin and Fermion Path Integrals
## 21.4 Summary