This inclusion on the wiki article occurred without any citations. It is not wrong, per se. If sharpness is 0 (resolution equal to uncertainty), then AUC is 0 or 1. If sharpness (refinement) is equal to uncertainty (no resolution), then the AUC is 0.5.
Other than that, it is not possible to further relate both quantities.
I tried to bring the formulas together below to identify perhaps boundaries that could be expressed in terms of either without success. I believe my discrete version of the AUC is not correct now (after some edits), but I would still like to offer it here for the insights that may already be visiblelet me know if anything is not clear.
Sharpness can be expressed as, assuming a number $K$ of discrete probabilities are issued, as:
\begin{align} \text{Sharpness} &= E[(X-E[X|P])^2] = \frac{1}{N}\sum_{i=1}^{K} \sum_{j=1}^{N_i} (x_{ij} - \bar x_i)^2\\ &= \frac{1}{N}\sum_{i=1}^{K} {N_i} \frac{N_i^+}{N_i} (1 - \bar x_i)^2 + {N_i}\frac{N_i^-}{N_i} \bar x_i^2 \frac{1}{N}\sum_{i=1}^{K} {N_i} \bar x_i (1 - \bar x_i)^2 + {N_i}(1 - \bar x_i) \bar x_i^2 = \\ &= \frac{1}{N}\sum_{i=1}^{K} {N_i} \bar x_i (1 - \bar x_i) \end{align}
$\bar x_i$ is the expected output in a given probability forecast, $x_{ij}$ represents the real outcome in example $j$ of the $i$-th probability forecast level. ${N_i}$ is the number of examples that are issued the $i$-th probability forecast level, and ${N_i}^+$ and ${N_i}^-$ are respectively the number of positive and negative examples at this particular forecast.
We can show that (see this answer)
\begin{align} \text{AUC} &= \int_0^1 \operatorname{TPR}(\operatorname{FPR}) d\operatorname{FPR}\\ &= \int_0^1 \operatorname{TPR}(\operatorname{FPR}(\tau)) d\operatorname{FPR}(\tau) \\ &= \int_{+\infty}^{-\infty} \operatorname{TPR}(\tau) \operatorname{FPR}'(\tau) d\tau \\ &= \int_{+\infty}^{-\infty} \big( 1-F_1(\tau) \big) \big( -f_0(\tau) \big) d\tau \\ &= \int_{-\infty}^{+\infty} \big( 1-F_1(\tau) \big) f_0(\tau) d\tau \\ &= \int_{-\infty}^{+\infty} f_0(\tau) d\tau - \int_{-\infty}^{+\infty} F_1(\tau) f_0(\tau) d\tau \\ &= 1 - \int_{-\infty}^{+\infty} \int_{-\infty}^{\tau}f_1(\tau') f_0(\tau) d\tau' d\tau \\ \end{align}\begin{align} \text{AUC} &= \int_0^1 \operatorname{TPR}(\operatorname{FPR}) d\operatorname{FPR}\\ &= \int_0^1 \operatorname{TPR}(\operatorname{FPR}(\tau)) d\operatorname{FPR}(\tau) \\ &= \int_{+\infty}^{-\infty} \operatorname{TPR}(\tau) \operatorname{FPR}'(\tau) d\tau \\ &= \int_{+\infty}^{-\infty} \big( 1-F_1(\tau) \big) \big( -f_0(\tau) \big) d\tau \\ &= \int_{-\infty}^{+\infty} \big( 1-F_1(\tau) \big) f_0(\tau) d\tau \\ &= \int_{-\infty}^{+\infty} f_0(\tau) d\tau - \int_{-\infty}^{+\infty} F_1(\tau) f_0(\tau) d\tau \\ &= 1 - \int_{-\infty}^{+\infty} \int_{-\infty}^{\tau}f_1(\tau') f_0(\tau) d\tau' d\tau \\ &= 1 - \int_{-\infty}^{+\infty} f_0(\tau) \int_{-\infty}^{\tau}f_1(\tau') d\tau' d\tau \\ \end{align}
Assuming again discrete probabilities for the AUC, we can write
\begin{align} \text{AUC} &= 1 - \int_{-\infty}^{+\infty} f_0(\tau) \int_{-\infty}^{\tau}f_1(\tau') d\tau' d\tau \rightarrow 1 - \frac{1}{N^-N^+}\sum_{i=1}^{K} N_i^- \sum_{j=1}^{i} N_j^+\\ &=1 - \frac{1}{N^2 \bar x (1 - \bar x)}\sum_{i=1}^{K} N_i \frac{N_i^+}{N_i} \sum_{j=1}^{i} N_j \frac{N_j^-}{N_j} = \\ &=1 - \frac{1}{N^2 \bar x (1 - \bar x)}\sum_{i=1}^{K}\sum_{j=1}^{i} N_i N_j \bar x_i (1 - \bar x_j) = \\ &=1 - \sum_{i=1}^{K}\sum_{j=1}^{i} \frac{N_i N_j}{N^2} \frac{\bar x_i (1 - \bar x_j)}{\bar x (1 - \bar x)} = \\ &=1 - \sum_{i=1}^{K} \frac{N_i^2}{N^2} \frac{\bar x_i (1 - \bar x_i)}{\bar x (1 - \bar x)} - \sum_{i=2}^{K}\sum_{j=1}^{i-1} \frac{N_i N_j}{N^2} \frac{\bar x_i (1 - \bar x_j)}{\bar x (1 - \bar x)}\\ \end{align}$$ \begin{align} \text{AUC} &= 1 - \frac{1}{N^-N^+}\sum_{i=1}^{K} N_i^- \sum_{j=1}^{i} N_j^+\\ &=1 - \frac{1}{N^2 \bar x (1 - \bar x)}\sum_{i=1}^{K} N_i \frac{N_i^-}{N_i} \sum_{j=1}^{i} N_j \frac{N_j^+}{N_j} = \\ &=1 - \frac{1}{N^2 \bar x (1 - \bar x)}\sum_{i=1}^{K}\sum_{j=1}^{i} N_i N_j \bar x_j (1 - \bar x_i) = \\ &=1 - \sum_{i=1}^{K}\sum_{j=1}^{i} \frac{N_i N_j}{N^2} \frac{\bar x_j (1 - \bar x_i) }{\bar x (1 - \bar x)} = \\ &=1 - \sum_{i=1}^{K} \frac{N_i^2}{N^2} \frac{\bar x_i (1 - \bar x_i)}{\bar x (1 - \bar x)} - \sum_{i=2}^{K}\sum_{j=1}^{i-1} \frac{N_i N_j}{N^2} \frac{\bar x_j (1 - \bar x_i) }{\bar x (1 - \bar x)}\\ \end{align} $$
If we assume $\bar x_i=\bar x$
$$ \begin{align} \text{AUC} &=1 - \sum_{i=1}^{K}\sum_{j=1}^{i} \frac{N_i N_j}{N^2} \frac{\bar x (1 - \bar x)}{\bar x (1 - \bar x)}\\ &=1 - \sum_{i=1}^{K}\sum_{j=1}^{i} \frac{N_i N_j}{N^2}\\ \end{align} $$
We can show that
$$N^2=\left(\sum_{i=1}^{K}N_i\right)^2=2\sum_{i=1}^{K}\sum_{j=1}^{i}N_iN_j$$
Thus
$$ \begin{align} \text{AUC} &=1 - \sum_{i=1}^{K}\sum_{j=1}^{i} \frac{N_i N_j}{N^2}\\ \end{align} =1-\frac{1}{2}\frac{N^2}{N^2}=0.5, $$
and $$ \begin{align} \text{Sharpness} &= \frac{1}{N}\sum_{i=1}^{K} {N_i} \bar x_i (1 - \bar x_i) = \frac{1}{N}\sum_{i=1}^{K} {N_i} \bar x (1 - \bar x)\\ &= \bar x (1 - \bar x)\sum_{i=1}^{K} \frac{N_i}{N}= \bar x (1 - \bar x)=\text{Uncertainty} \end{align} $$
If we assume $\bar x_i \in \{0,1\}$
$$ \begin{align} \text{Sharpness} &= \frac{1}{N}\sum_{i=1}^{K} {N_i} \bar x_i (1 - \bar x_i) = 0 \end{align} $$
This is the perfect Sharpness score, where the Uncertainty is negated by the forecast Resolution. On the other hand,
$$ \begin{align} \text{AUC} &=1 - \sum_{i=1}^{K}\sum_{j=1}^{i} \frac{N_i N_j}{N^2} \frac{\bar x_i (1 - \bar x_j)}{\bar x (1 - \bar x)}\\ &=1 - \sum_{i=2}^{K}\sum_{j=1}^{i-1} \frac{N_i N_j}{N^2} \frac{\bar x_i (1 - \bar x_j)}{\bar x (1 - \bar x)}\\ \end{align} $$$$ \begin{align} \text{AUC} &=1 - \sum_{i=1}^{K}\sum_{j=1}^{i} \frac{N_i N_j}{N^2} \frac{\bar x_j (1 - \bar x_i) }{\bar x (1 - \bar x)}\\ &=1 - \sum_{i=2}^{K}\sum_{j=1}^{i-1} \frac{N_i N_j}{N^2} \frac{\bar x_j (1 - \bar x_i) }{\bar x (1 - \bar x)}\\ \end{align} $$
Here, we have to further assume that there is an index $i'$ for which $\bar x_{i}=1 ~\forall~ i>i', 1\lt i' \lt K$, $\bar x_i = 0$ otherwise. This is where the intrinsic ordering of the scores shows up in the perfect AUC score.
With that assumption, we can split the sums, once for each index, $$ \begin{align} \text{AUC} &=1 - \overbrace{\sum_{i=2}^{i'}\sum_{j=1}^{i-1} \frac{N_i N_j}{N^2} \frac{0 \cdot (1 - \bar x_j)}{\bar x (1 - \bar x)}}^{=0~\because~ \bar x_i=0~\forall~i\leq i'} - \sum_{i=i'+1}^{K}\sum_{j=1}^{i-1} \frac{N_i N_j}{N^2} \frac{1 \cdot (1 - \bar x_j)}{\bar x (1 - \bar x)}\\ &=1 - \sum_{i=i'+1}^{K}\sum_{j=1}^{i'} \frac{N_i N_j}{N^2} \frac{(1 - \overbrace{0}^{\bar x_j=0~\because~j\leq~i'})}{\bar x (1 - \bar x)} - \sum_{i=i'+2}^{K}\sum_{j=i'+1}^{i-1} \frac{N_i N_j}{N^2} \frac{(1 - \overbrace{1}^{\bar x_j=1~\because~j\gt~i'})}{\bar x (1 - \bar x)}\\ &=1 - \sum_{i=i'+1}^{K}\sum_{j=1}^{i'} \frac{N_i N_j}{N^2} \frac{1}{\bar x (1 - \bar x)} =1 - \frac{\sum_{i=i'+1}^{K} N_i \sum_{j=1}^{i'}N_j}{N^2 \bar x (1 - \bar x)}\\ \end{align} $$
$$ \begin{align} \text{AUC} &=1 - \sum_{i=2}^{i'}\sum_{j=1}^{i-1} \frac{N_i N_j}{N^2} \frac{\bar x_j \cdot (1 - \overbrace{0}^{\bar x_i=0~\forall~i\leq i'})}{\bar x (1 - \bar x)} - \overbrace{\sum_{i=i'+1}^{K}\sum_{j=1}^{i-1} \frac{N_i N_j}{N^2} \frac{\bar x_j \cdot (1 - 1)}{\bar x (1 - \bar x)}}^{=0~\because~ \bar x_i=1~\forall~i\gt i'}\\ &=1 - \sum_{i=2}^{i'}\sum_{j=1}^{i-1} \frac{N_i N_j}{N^2} \frac{\overbrace{\bar x_j}^{=0~\because~j<i\leq i'}}{\bar x (1 - \bar x)}=1 \end{align} $$
Similarly, assuming instead that here is an index $i'$ for which $\bar x_{i}=1 ~\forall~ i\leq i', 1\lt i' \lt K$, $\bar x_i = 0$ otherwise, we once again can split the sums,
$$ \begin{align} \text{AUC} &=1 - \overbrace{\sum_{i=2}^{i'}\sum_{j=1}^{i-1} \frac{N_i N_j}{N^2} \frac{\bar x_j \cdot (1 - 1)}{\bar x (1 - \bar x)}}^{=0~\because~ \bar x_i=1~\forall~i\leq i'} - \sum_{i=i'+1}^{K}\sum_{j=1}^{i-1} \frac{N_i N_j}{N^2} \frac{\bar x_j \cdot (1 - 0)}{\bar x (1 - \bar x)}\\ &=1 - \sum_{i=i'+1}^{K}\sum_{j=1}^{i'} \frac{N_i N_j}{N^2} \frac{\overbrace{(1) }^{\bar x_j=1~\because~j\leq~i'}}{\bar x (1 - \bar x)} - \sum_{i=i'+2}^{K}\sum_{j=i'+1}^{i-1} \frac{N_i N_j}{N^2} \frac{\overbrace{(0)}^{\bar x_j=0~\because~j\gt~i'}}{\bar x (1 - \bar x)}\\ &=1 - \sum_{i=i'+1}^{K}\sum_{j=1}^{i'} \frac{N_i N_j}{N^2} \frac{1}{\bar x (1 - \bar x)} =1 - \frac{\left(\sum_{i=i'+1}^{K} N_i\right) \left(\sum_{j=1}^{i'}N_j\right)}{N^2 \bar x (1 - \bar x)}\\ \end{align} $$
Due to the intrinsic ordering we introduced
$$\sum_{i=i'+1}^{K}N_i=\bar xN$$
is the total number of positives examples and, similarly,
$$\sum_{j=1}^{i'}N_j=(1-\bar x)N$$
is the sum of all negative examples. Thus
$$ \begin{align} \text{AUC} &=1 - \frac{\sum_{i=i'+1}^{K} N_i \sum_{j=1}^{i'}N_j}{N^2 \bar x (1 - \bar x)}\\ &=1-\frac{N^2\bar x(1-\bar x)}{N^2 \bar x (1 - \bar x)}=0 \end{align} $$
Similarly, assuming instead that here is an index $i'$ for which $\bar x_{i}=1 ~\forall~ i\leq i', 1\lt i' \lt K$, $\bar x_i = 0$ otherwise, we once again can split the sums, $$ \begin{align} \text{AUC} &=1 - \sum_{i=2}^{i'}\sum_{j=1}^{i-1} \frac{N_i N_j}{N^2} \frac{1 \cdot (1 - \bar x_j)}{\bar x (1 - \bar x)}- \overbrace{\sum_{i=i'+1}^{K}\sum_{j=1}^{i-1} \frac{N_i N_j}{N^2} \frac{0 \cdot (1 - \bar x_j)}{\bar x (1 - \bar x)}}^{=0~\because~ \bar x_i=0~\forall~i\geq i'}\\ &=1 - \sum_{i=2}^{i'}\sum_{j=1}^{i-1} \frac{N_i N_j}{N^2} \frac{(1 - \overbrace{1}^{\bar x_j=1~\because~j\lt i \leq~i'})}{\bar x (1 - \bar x)} \\ &=1 - \sum_{i=i'+1}^{K}\sum_{j=1}^{i'} \frac{N_i N_j}{N^2} \frac{1}{\bar x (1 - \bar x)} =1\\ \end{align} $$$$ \begin{align} \text{AUC} &=1 - \frac{\left(\sum_{i=i'+1}^{K} N_i\right) \left(\sum_{j=1}^{i'}N_j\right)}{N^2 \bar x (1 - \bar x)} &=1-\frac{N^2\bar x(1-\bar x)}{N^2 \bar x (1 - \bar x)}=0 \end{align} $$
This way we show that perfectly separated classes per bin will result in $\text{Sharpness}=0$ and $\text{AUC}=0$ or $\text{AUC}=1$, depending on the orientation of the scores.