# Baby Rudin theorem 10.33 ( STOKES' THEOREM)

The definitions which we need for the proof of the theorem. We define the standard simplex $$Q^k$$ to be the set of all $$u$$ $$\in$$ $$R^k$$ of the form $$u$$ = $$\sum_{i=1}^k$$ $$\alpha_i$$ $$e_i$$.

Assume now that $$p_0$$, $$p_1$$,...$$p_k$$ are points of $$R^n$$.

The oriented affine $$k$$-simplex $$\sigma$$ $$=$$ [$$p_0$$, $$p_1$$,...$$p_k$$] is defined to be the $$k$$-surface in $$R^n$$ with parameter domain $$Q^k$$ which is given by the affine mapping $$\sigma$$($$\sum_{i=1}^k$$ $$\alpha_i$$ $$e_i$$) $$=$$ $$\sigma(u)$$ $$=p_0$$ + $$\sum_{i=1}^k$$ $$\alpha_i$$($$p_i$$ - $$p_0$$). Note that $$\sigma$$ is characterized by $$\sigma(0)$$ = $$p_0$$, $$\sigma(e_i)$$ = $$p_i$$ (for $$1$$ $$\leq$$ $$i$$ $$\leq$$ $$k$$).

For $$k$$ $$\geq$$ $$1$$, the boundary of the oriented affine $$k$$-simplex

$$\sigma$$ $$=$$ [$$p_0$$, $$p_1$$,...$$p_k$$] is defined to be the affine ($$k-1$$)-chain

$$\partial\sigma$$ = $$\sum_{j=0}^k$$ $$(-1)^j$$[$$p_0$$,...,$$p_{j-1}$$,$$p_{j+1}$$,..,$$p_k$$]. For $$1$$ $$\leq$$ $$j$$ $$\leq$$ $$k$$, observe that the simplex $$\sigma_j$$ = [$$p_0$$,...,$$p_{j-1}$$,$$p_{j+1}$$,..,$$p_k$$] has $$Q^{k-1}$$ as its parameter domain and that is defined by

$$\sigma_j(u)$$ = $$p_0$$ + $$Bu$$ ( $$u$$ $$\in$$ $$Q^{k-1}$$)

The class $$\mathscr C'$$ means the class of continuously differentiable functions and etc.

$$x_j$$ = $${ \begin{cases} {u_j (1 \leq j \lt r),} \\ {1 - (u_1 + ... + u_{k-1}) (j=r),} \\ {u_{j-1} (r \lt j \leq k). } \end{cases} }$$

( this is $$(98)$$).

$$x_j$$ = $${ \begin{cases} {u_j (1 \leq j \lt i),} \\ { 0 (j=i),} \\ {u_{j-1} (i \lt j \leq k). } \end{cases} }$$ .

( this is $$(99)$$).

I dont'understand how do we get the $$(98)$$ and $$(99)$$.

Any help would be appreciated.

• Maybe you should write what is $Q^{k-1}$ and how $[e_1,e_2,\dots,e_k](\mathbf{u})$ ist defined as a map. Commented Mar 10, 2022 at 10:40
• @hal4math Check it. I did it. Commented Mar 10, 2022 at 11:26

You have $$x = \tau_0(u) = e_r + \sum_{i=1}^{k-1} u_i(p_i - e_r)$$. So if $$1 \leq j < r$$, then you get

$$x_j = \left(e_r + \sum_{i=1}^{k-1} u_i p_i -\sum_{i=1}u_i e_r\right)_j = u_j,$$

since for $$j < r$$ we have $$p_j = e_j$$. However, for $$j > r$$ we have $$p_{j-1} = e_{j}$$, so that in that case

$$x_j = \left(e_r + \sum_{i=1}^{k-1} u_i p_i -\sum_{i=1}u_i e_r\right)_j = u_{j-1}.$$

For the case $$j=r$$, it is

\begin{aligned} x_j &= \left(e_r + \sum_{i=1}^{k-1} u_i p_i -\sum_{i=1}^{k-1} u_i e_r\right)_j \\ &= \left(e_r + \sum_{i=1}^{k-1} u_i p_i -\sum_{i=1}^{k-1} u_i e_r\right)_j \\ &= \left(e_r -\sum_{i=1}^{k-1} u_i e_r\right)_j \\ &= \left((1 -\sum_{i=1}^{k-1} u_i) e_r\right)_j \\ &= 1 -\sum_{i=1}^{k-1} u_i. \end{aligned}

Hope that helps!

• sorry I don't understand how did you get $u_j$ in the fist line. please explain it more explicitly. Commented Mar 11, 2022 at 11:34
• You have $\sum_{i=1}^{k-1} u_i p_i = u_1p_1 + \dots u_{r-1}p_{r-1} + \dots u_{r}p_{r} +\dots u_{k-1}p_{k-1}$. So plugging in the $p_i$s you get $u_1e_1 + \dots u_{r-1}e_{r-1} + \dots + u_{r}e_{r+1} +\dots + u_{k-1}e_{k}$. There you see that the $j$-th component is $u_j$. Notice that I should not have summed up to $k$ but only $k-1$. Commented Mar 11, 2022 at 11:47
• the j-th component is $u_j$$e_j$ . and why are the other other components equal of zero? @hal4math Commented Mar 11, 2022 at 11:49
• They are not zero, but they are not in the j-th component? Also the $j$-th component is $u_j$ and not $u_j e_j$. That is still a vector. Commented Mar 11, 2022 at 11:51
• okay I get it. but could you tell why is it still a vector? Commented Mar 11, 2022 at 12:04