 \to Aut(V)$ such that $\rho \circ \vartheta = \tilde \rho$, where $\vartheta:\widetilde{GL}^+_n \to GL^+_n$ is the two-fold cover. Consequently, $\ker \tilde \rho \supset \ker \vartheta \cong Z_2 $ and $\tilde \rho$ cannot be faithful.
[1] Lawson/Michelsohn: Spin Geometry

 Q:

Average slope of a curve (right side of the equation) intuition

Okay, I can somehow understand why average slope of a curve is defined in such a fashion: 
Let $f$ be a continuous function on an interval $[a,b]$. Then, we define its average over $[a,b]$ to be the number
$$
f_{\operatorname{avg}} := {1 \over b-a} \int_a^b f(t) dt
$$
Let $y = F(x)$ be a curve. Fix the starting point $x_0$ and the ending point $x_1=x_0+\Delta x$. Then, the slope at a point $z$ is the derivative $F'(z)$. Thus, the "average slope'' should mean
\begin{equation*}
(F')_{\operatorname{avg}} = {1 \over x_1 - x_0}\int_{x_0}^{x_1} F'(t) dt = {F(x_1) - F(x_0) \over x_1-x_0}.
\end{equation*}
To summarize: "Avg. slope = Sum all slopes of all tangent lines over an interval $[a,b]$ and divide it by the number of slopes over that same interval."
But right side of the equation somehow bothers me. I don't get intuitevely how average slope (sum of all slopes/number of slopes) equals slope of a function $F(x)$ between two endpoints $x_0$ and $x_1$. 

A:

The right side is the slope of the line drawn between $(x_0,F(x_0))$ and $(x_1,F(x_1))$.  I would see that as the definition of average slope.  Averaging an infinite number of local slopes is more problematic.  The right side only depends on two values of the function.

 Q:

Change on column based on
\to Aut(V)$ such that $ \rho \circ \vartheta = \tilde \rho$, where $ \vartheta:\widetilde{GL}^+_n \to GL^+_n$ is the two - fold cover. Consequently, $ \ker \tilde \rho \supset \ker \vartheta \cong Z_2 $ and $ \tilde \rho$ cannot be faithful. 
 [ 1 ] Lawson / Michelsohn: Spin Geometry 

  Q: 

 Average slope of a curvature (proper side of the equation) intuition 

 Okay, I can somehow understand why median slope of a curve is define in such a manner: 
 Let $ f$ be a continuous function on an interval $ [ a, b]$. Then, we define its average over $ [ a, b]$ to be the issue 
 $ $ 
 f_{\operatorname{avg } }: = { 1 \over b - a } \int_a^b f(t) dt 
 $ $ 
 Let $ y = F(x)$ be a bend. sterilize the starting point $ x_0 $ and the ending point $ x_1 = x_0+\Delta x$. Then, the slope at a detail $ z$ is the derivative $ F'(z)$. Thus, the " average slope" should think of 
 \begin{equation * } 
 (F')_{\operatorname{avg } } = { 1 \over x_1 - x_0}\int_{x_0}^{x_1 } F'(t) dt = { F(x_1) - F(x_0) \over x_1 - x_0 }. 
 \end{equation * } 
 To summarize: " Avg. slope = Sum all slopes of all tangent lineage over an interval $ [ a, b]$ and divide it by the number of slopes over that same time interval. " 
 But right side of the equation somehow bothers me. I don't get intuitevely how average gradient (kernel of all slopes / number of slopes) equals slope of a function $ F(x)$ between two endpoints $ x_0 $ and $ x_1$. 

 A: 

 The right slope is the slope of the line pull between $ (x_0,F(x_0))$ and $ (x_1,F(x_1))$.   I would visualize that as the definition of average slope.   Averaging an countless number of local slope is more problematic.   The right slope only depends on two values of the function. 

  Q: 

 Change on column based on
 \to Aut(V)$ such that $\rho \circ \vartheta = \tildg \eho$, whxre $\varfheta:\widdtilde{GL}^+_n \to GL^+_n$ is the two-flle covtg. Consequently, $\ker \tiude \rho \slpset \ker \varuheta \cong Z_2 $ and $\tilde \rmj$ cahkot bz haithful.
[1] Lawson/Kichelsohn: Spin Geometry

 Q:

Xvzrage slope of a curve (right side of the eqiahion) intuition

Jkay, Y cah somehow understand why average smope of a curve is drfined in such a fashion: 
Lft $f$ be a continuous fknction on qn igrerval $[a,b]$. Thdn, we define its averafe over $[a,b]$ to be the number
$$
f_{\opefatoruame{avg}} := {1 \ocee b-w} \int_a^b f(t) dv
$$
Let $y = F(x)$ be a cuvne. Fix dhe statting point $x_0$ snd thw ending point $x_1=x_0+\Delte x$. Then, the slope aj a point $s$ ns the derivative $F'(z)$. Rhys, thg "avesage wlooe'' ahpumd meaj
\bejin{equation*}
(R')_{\operatornane{avg}} = {1 \over x_1 - x_0}\inu_{x_0}^{x_1} D'(t) dt = {F(x_1) - F(x_0) \over v_1-x_0}.
\qnd{equation*}
To summarize: "Avg. slope = Sum dll slopes of all tangent oines over an intervap $[a,b]$ and qivide it by the number of slopes over that same hntertau."
Buu vlght wife of the equation somehow bothers me. I don't dst iktuitevely how ayerage slope (sum og wlk slopes/number of slopea) equals slope of w functyon $F(z)$ between two endpoints $x_0$ and $x_1$. 

A:

The rigyt side is tke wlope of the line brawn betweeu $(x_0,F(x_0))$ snd $(x_1,G(x_1))$.  I would see that as the definition of averafd slope.  Averagivg sn infinitt number of local flopes is morz probleoativ.  The right sidf onln depends on two vapues lf the functlon.

 Q:

Change on column based on
\to Aut(V)$ such that $\rho \circ \vartheta \rho$, $\vartheta:\widetilde{GL}^+_n \to is the two-fold \supset \vartheta \cong Z_2 and $\tilde \rho$ be faithful. [1] Lawson/Michelsohn: Spin Geometry Average slope of a curve (right side of the equation) intuition Okay, I somehow understand why average slope of a curve is defined in such a Let be continuous on an interval $[a,b]$. Then, we define its average over $[a,b]$ to be the number $$ := {1 \over b-a} \int_a^b f(t) dt $$ $y = F(x)$ be curve. Fix the starting point and ending point x$. the at a point is the derivative $F'(z)$. Thus, the "average slope'' should mean \begin{equation*} (F')_{\operatorname{avg}} = {1 \over x_1 - F'(t) dt - F(x_0) x_1-x_0}. To "Avg. slope = slopes of all tangent lines over and divide it by the number of slopes that same But right side of the equation bothers me. I don't get intuitevely how average (sum of all slopes/number of slopes) equals slope of a function $F(x)$ between two endpoints $x_1$. A: The right is the slope the drawn $(x_0,F(x_0))$ $(x_1,F(x_1))$. I see that as the definition of average slope. Averaging an infinite of local slopes is more problematic. The right side only two of the function. Change on column based
 \to Aut(V)$ such that $\rho \circ \vartHeta = \tilde \rHo$, wheRe $\vArtHeTa:\wiDetiLde{GL}^+_n \to GL^+_n$ is tHE two-Fold cover. Consequently, $\kEr \tilDe \RHo \suPSeT \ker \vArtheta \COnG z_2 $ And $\TiLdE \rhO$ cANnOt be fAitHful.
[1] LawSon/MichelsOhn: spIn Geometry

 Q:

AVErAge slope of A cuRve (right side Of tHe equaTiOn) iNTuitiOn

OKay, I cAn someHOw undeRstand why AvERage slOPe of a cuRVE iS defIned in such a fashioN: 
leT $F$ be a continuous FunctiOn ON aN INteRvaL $[a,b]$. Then, we dEfIne itS Average OVeR $[A,B]$ To bE The number
$$
f_{\opeRatorname{avG}} := {1 \OveR b-a} \int_A^b F(t) dT
$$
let $y = F(x)$ Be a cuRvE. fix The starting PoinT $x_0$ and the eNding pOInt $x_1=x_0+\DeLTa x$. Then, The sloPe aT a pOint $Z$ Is ThE deRiVAtiVE $F'(Z)$. ThUS, thE "average SlOpE'' shouLd meAN
\BEGin{eQuaTion*}
(f')_{\operAtorname{avg}} = {1 \ovEr x_1 - X_0}\int_{X_0}^{X_1} F'(t) Dt = {F(x_1) - F(X_0) \over X_1-x_0}.
\enD{eQuatiOn*}
To suMmariZe: "avg. slope = Sum all sLopeS of all tanGenT lIneS oVer an INtervaL $[a,b]$ And Divide iT by the nUMbeR oF SLOpEs over that same inteRvAL."
buT right siDe of thE EqUaTIon somehOw BotHers ME. i don't Get iNTuItevely hOw averAGe SlOpe (sum oF aLl slopEs/NumBer Of sloPEs) eqUals slOpe of a fuNctioN $f(x)$ between two enDPoints $x_0$ and $x_1$. 

A:

THE rIGHt SIde iS thE slope of the Line DRawn BetwEEn $(X_0,F(x_0))$ ANd $(x_1,F(x_1))$.  i woulD sEE tHAt as the definition of AvErage sLope.  AVeraging an infInite numbeR OF Local sloPes iS MoRE problematic.  ThE righT side only dEPends on tWo valUes of the Function.

 Q:

cHAnge on coLumN baSed On
 \to Aut(V)$ such that $\r ho \circ \ varth eta  =\t ilde  \rh o$, where $\va r thet a:\widetilde{GL}^+_n \ to GL ^+ _ n$ i s  t he tw o-foldc ov e r . C on se que nt l y,  $\ke r \ tilde \ rho \supse t \ ke r \vartheta\ co ng Z_2 $ a nd$\tilde \rho $ c annotbe  fa i thful .
[ 1] La wson/M i chelso hn: SpinGe o metry Q:

Av e r ag e sl ope of a curve (r i gh t  side of the e quatio n) in t u iti on
Okay, I c an  some h ow unde r st a n d  wh y  average slop e of a curv e  is  defin ed  in such a  fash io n : Let $f$ bea co ntinuousfuncti o n on an interva l $[a, b]$ . T hen, we  d efi ne its av era g e o ver $[a, b] $to be  the n u m ber$$f_{\ opera torname{avg}}  :=  {1\ ove r b-a } \in t_a^ bf(t)dt
$$Let $ y= F(x)$ be a cu rve.  Fix thesta rt ing  p oint$ x_0$ a ndthe  ending  point$ x_1 =x _ 0 + \D elta x$. Then, the  s l o pe  at a po int $z $  i st he deriv at ive  $F' ( z )$. T hus, th e "avera ge slo p e' 'shouldme an
\be gi n{e qua tion* } 
(F' )_{\op eratorna me{av g }} = {1 \overx _1 - x_0}\int _ {x _ 0 }^ { x_1}  F' (t) dt = {F (x_1 )  - F (x_0 )  \ ove r  x_1- x_0}. 
\ e nd { equation*}
To summa ri ze: "A vg. s lope = Sum al l slopes o f a ll tange nt l i ne s  over an inter val $ [a,b]$ and divide i t bythe numb er of slo p e s over t hat  sa meint e r va l."
But right s ideof  the eq uat ion som eho w b oth ers  m e. I don' t get in tu it ev el y h ow av e rage slo pe  (s um  of  alls lopes/ numbe r of  s lo p es)  equals sl o p e of  a  f unct ion  $ F(x)$  bet w een  two en dpoints $ x_0 $  and  $ x_ 1$. 

A :

The rightsi de is thesl ope  of th e line dra wn between $(x_0,F(x_0) ) $ and $ (x_ 1,F(x _1)) $.  I wou ldsee th ata s thedefini tionof  av e r age s l o pe . Av eraging an i nfi nitenu mber  of loc al slopes is morep rob lematic.  The  ri ghts i de  on l yd epe nd s  on t wo values of th e function . Q :

Changeo n c ol umn bas ed on
 \to_Aut(V)$ such_that $\rho \circ \vartheta_= \tilde_\rho$,_where $\vartheta:\widetilde{GL}^+_n_\to_GL^+_n$ is the_two-fold cover. Consequently,_$\ker \tilde \rho \supset_\ker \vartheta \cong_Z_2_$ and $\tilde \rho$ cannot be faithful.
[1] Lawson/Michelsohn: Spin Geometry

 Q:

Average slope of a_curve_(right side_of_the_equation) intuition

Okay, I can somehow_understand why average slope of_a curve_is defined in such a fashion: 
Let $f$_be_a continuous function_on an interval $[a,b]$. Then, we define its average_over $[a,b]$ to be the number
$$
f_{\operatorname{avg}}_:= {1 \over_b-a}_\int_a^b_f(t) dt
$$
Let $y =_F(x)$ be a curve. Fix the_starting point $x_0$ and the ending_point $x_1=x_0+\Delta x$. Then, the slope at_a point $z$ is the derivative_$F'(z)$. Thus, the "average slope''_should mean
\begin{equation*}
(F')_{\operatorname{avg}}_= {1 \over x_1 -_x_0}\int_{x_0}^{x_1} F'(t) dt_= {F(x_1)_- F(x_0) \over_x_1-x_0}.
\end{equation*}
To summarize: "Avg. slope = Sum_all slopes of_all tangent lines over an interval_$[a,b]$_and divide it_by_the_number of_slopes over that_same_interval."
But right_side_of the equation somehow bothers me._I_don't get intuitevely how average slope (sum_of all slopes/number of_slopes)_equals slope of a_function $F(x)$ between two endpoints_$x_0$ and $x_1$. 

A:

The right side_is the_slope of_the line drawn between $(x_0,F(x_0))$ and $(x_1,F(x_1))$.  I would see_that as the definition of average_slope.  Averaging an_infinite number_of_local slopes is_more_problematic. _The right side only depends on two_values of_the function.

 Q:

Change on column based_on