Accents 包与 bm 不兼容吗？

Question

阅读埃格尔的回答减少数学下划线的长度，我想使用accents数学模式下较低条形的包。

然而，该软件包似乎与bm包。当我添加\usepackage{accents}到我的序言中时，即使没有使用与包相关的任何命令，我也会收到所有使用 的错误消息\bm。

Missing number, treated as zero ...s when only one specific $\bm{\tilde{a}}

Missing control sequence inserted ...s when only one specific $\bm{\tilde{a}}

Missing } inserted ...s when only one specific $\bm{\tilde{a}}

Missing } inserted ...s when only one specific $\bm{\tilde{a}}

Missing } inserted ...s when only one specific $\bm{\tilde{a}}

Missing } inserted ...s when only one specific $\bm{\tilde{a}}

Extra }, or forgotten \endgroup ...s when only one specific $\bm{\tilde{a}}

Missing control sequence inserted ...s when only one specific $\bm{\tilde{a}}

Parameters must be numbered consecutively ...s when only one specific $\bm{\tilde{a}}

Missing number, treated as zero ... \leq \left| \sum_{a_i \in \bm{\tilde{a}}

Missing control sequence inserted ... \leq \left| \sum_{a_i \in \bm{\tilde{a}}

Missing } inserted ... \leq \left| \sum_{a_i \in \bm{\tilde{a}}

Extra }, or forgotten \endgroup ... \leq \left| \sum_{a_i \in \bm{\tilde{a}}

Missing control sequence inserted ... \leq \left| \sum_{a_i \in \bm{\tilde{a}}

Parameters must be numbered consecutively ... \leq \left| \sum_{a_i \in \bm{\tilde{a}}

Missing number, treated as zero for any $\bm{\tilde{a}}

Missing control sequence inserted for any $\bm{\tilde{a}}

Missing } inserted for any $\bm{\tilde{a}}

Extra }, or forgotten \endgroup for any $\bm{\tilde{a}}

Missing control sequence inserted for any $\bm{\tilde{a}}

Parameters must be numbered consecutively for any $\bm{\tilde{a}}

Missing number, treated as zero ...he case in particular for $\bm{\tilde{a}}

Missing control sequence inserted ...he case in particular for $\bm{\tilde{a}}

Missing } inserted ...he case in particular for $\bm{\tilde{a}}

Extra }, or forgotten \endgroup ...he case in particular for $\bm{\tilde{a}}

Missing control sequence inserted ...he case in particular for $\bm{\tilde{a}}

Parameters must be numbered consecutively ...he case in particular for $\bm{\tilde{a}}

有没有办法使这两个包兼容？

Answer 1

而要$\bm{\tilde{a}}$使用$\tilde{\bm{a}}$。

或者

使用$\bm{{\tilde{a}}}$，里面有一个额外的分组\bm{{...}}。