Construction of the $E_8$ root system The following construction of $E_8$ root system is introduced in J.Humphreys' Introduction to Lie Algebras and Representation Theory(3rd printing, p.65). Some notations are slightly changed from the book.


*

*Take an orthonormal basis $e_1, e_2, \cdots, e_8$ of $E=\mathbb R^8$

*Consider the $\mathbb{Z}$-module $I$ generated by $e_1, e_2\cdots, e_8$ and $\delta = \frac{1}{2}(e_1+ \cdots + e_8)$.

*Let $I'$ be the subgroup of $I$ consisting of all elements $c\delta +\sum_{i=1}^8 c_i e_i $ $(c, c_i\in \mathbb Z$), for which $c + \sum_{i=1}^8 c_i$ is an even integer.

*Define $\Phi = \{ \alpha \in I' \colon (\alpha, \alpha)=2\}$. Calculation shows that $$\Phi =\{ \pm (e_i \pm e_j) \mid i \neq j\} \cup \{ \frac{1}{2} \sum_{i=1}^{8} (-1)^{k(i)}e_i \mid k(i) \in \{0,1\}, \sum k(i) \text{ is even}\}$$
It can be verified that $\Phi$ is indeed a root system, by showing $\langle \alpha, \beta \rangle \in \mathbb Z$ for all $\alpha, \beta \in I'$.

*$\Delta = \{\frac{1}{2}\left( e_1 + e_8 - (e_2 + \cdots + e_6) \right), \, e_1+e_2, \, e_2-e_1, \, \,e_3-e_2, \, e_4-e_3, \, e_5-e_4, \, e_6-e_5, \, e_7-e_6  \}$ is a base of $\Phi$.



I proved that $I'$ is well-defined, as follows:  
Claim: $c \delta + \sum_{i=1}^8 c_i e_i  = d\delta+ \sum_{i=1}^8 d_ie_i$ implies $ c + \sum c_i  = d+ \sum d_i \mod 2$.
proof: $c_i+ c/2 = d_i + d/2$ for all $i= 1, \dots, 8$. In particular, $c_1 - d_1 = \frac{1}{2} (d-c) \in \mathbb Z$ implies $d \equiv c$ $\text{mod}$ $2$. 
Thus $$ c+ \sum_{i=1}^8 c_i = c+\sum_{i=1}^8 (d_i+ \frac{d-c}{2}) = c+ \sum_{i=1}^{8}d_i +4(d-c) \equiv d+ \sum_{i=1}^{8}d_i \mod 2$$

But after this I noticed something strange above.  
First, $\alpha_1 :=\frac{1}{2}\left( e_1 + e_8 - (e_2 + \cdots + e_6) \right) \in \Delta \subset I'$ but $\alpha_1 = \delta  + \sum_{i=1}^{6} (-1)e_i $ implies $\alpha_1 \in I \setminus I'$. $(1-6=-5)$
Second, take $\alpha = \frac{1}{2} \sum_{i=1}^{8} (-1)^{k(i)}e_i \in \Phi$. Then $\alpha =\delta + \sum_{i=1}^8 \frac{ (-1)^{k(i)}-1 }{2} e_i $ so $ 1 + \sum_{i=1}^8\frac{ (-1)^{k(i)}-1 }{2}$ must be an even integer. But this number equals to $1 +\frac{1}{2}(-r + (8-r))-4 =1-r$ where $r$ is the number of $k(i)$ which equals to $1$. Thus $r$ should be odd, and the calculation of $\Phi $ given in 3. is not correct.
I couldn't find where I went wrong. Is my reasoning correct, or am I missing something?

EDIT
I misread $\frac{1}{2}\left( e_1 + e_8 - (e_2 + \cdots + e_7) \right)$ as a $\frac{1}{2}\left( e_1 + e_8 - (e_2 + \cdots + e_6) \right)$, so $\Delta$ equals to $\{\frac{1}{2}\left( e_1 + e_8 - (e_2 + \cdots + e_7) \right), \, e_1+e_2, \, e_2-e_1, \, \,e_3-e_2, \, e_4-e_3, \, e_5-e_4, \, e_6-e_5, \, e_7-e_6  \}$, accordingly.
The below image is the original text. 

 A: This is a slightly eccentric definition of the root lattice. Usually one would define
it as the vectors of the form $(a_1,\ldots,a_8)$ where the $a_i$ are all integers
or all half-integers, and the sum of the $a_i$ is even.
(By a "half-integer" I mean an integer plus $1/2$.)
This definition comes out as $I'$ being the set of
vectors of the form $(a_1,\ldots,a_8)$ where the $a_i$ are all integers
or all half-integers, and $-a_1+\sum_{i=2}^8a_i$ is even.
This doesn't matter, since it's the image of the standard root lattice under
the reflection $(a_1,\ldots,a_8)\mapsto(-a_1,a_2,\ldots,a_8)$.
But however you cut it, $\frac12(e_1+e_8-(e_2+\cdots+e_6))$ isn't in the
root lattice. Is that a misprint for $\frac12(e_1+e_8-(e_2+\cdots+e_7))$?
The roots in the lattice $I'$ will be the $112$ vectors $\pm e_1\pm e_j$
for $i<j$ and the $128$ vectors $\frac12\sum_1^8\pm e_i$ with an odd number
of minus signs.
As I say, it's more conventional to define the root lattice so that
it contains the vectors $\frac12\sum_1^8\pm e_i$ with an even number
of minus signs.
